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a.
\(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\left(x\ge-1\right)\)
\(B=\sqrt{16}.\sqrt{x+1}-\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}+\sqrt{x+1}\)
\(B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(B=\left(4-3+2+1\right).\sqrt{x+1}\)
\(B=4.\sqrt{x+1}\)
b.
\(B=16\\\)
\(\Rightarrow4\sqrt{x+1}=16\)
\(\Rightarrow\sqrt{x+1}=\dfrac{16}{4}=4\)
\(\Rightarrow x+1=4^2\)
\(\Rightarrow x+1=16\rightarrow x=16-1=15\) (thỏa mãn)
vậy x=15
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
a) \(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)
\(=\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}\)
\(=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=4\sqrt{x+1}\)
b) \(B=4\sqrt{x+1}=16\) khi \(\sqrt{x+1}=4\) hay x+1=16 => x=15
a) \sqrt{-9a}-\sqrt{9+12 a+4 a^{2}}−9a−9+12a+4a2
=\sqrt{-9 a}-\sqrt{3^{2}+2.3 .2 a+(2 a)^{2}}=−9a−32+2.3.2a+(2a)2
=\sqrt{3^{2} \cdot(-a)}-\sqrt{(3+2 a)^{2}}=32⋅(−a)−(3+2a)2
=3 \sqrt{-a}-|3+2 a|=3−a−∣3+2a∣
Thay a=-9a=−9 ta được:
3 \sqrt{9}-|3+2 \cdot(-9)|=3.3-15=-639−∣3+2⋅(−9)∣=3.3−15=−6.
b) Điều kiện: m \neq 2m=2
a) \(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}\)
\(=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}=-5\sqrt{x-1}\)
b) \(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)
\(=5\sqrt{y+4}+6\sqrt{y+4}-18\sqrt{y+4}=-7\sqrt{y+4}\)
c) \(P=\sqrt{y-2}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)
\(=\sqrt{y-2}-24\sqrt{y-2}+28\sqrt{y-2}=5\sqrt{y-2}\)
a) \(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}.\)
\(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}\)
\(=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}\)
\(=-5\sqrt{x-1}\)
b) \(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)
\(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)
\(=5\sqrt{y+4}+6\sqrt{y+4}\)
\(=-7\sqrt{y+4}\)
c) \(P=\sqrt{\left(y-2\right)}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)
\(P=\sqrt{\left(y-2\right)}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)
\(=\sqrt{y-2}-24\sqrt{y-2}+28\sqrt{y-2}\)
\(=5\sqrt{y-2}\)
Rút gọn các biểu thức sau với x≥0x≥0:
a) 2\(\sqrt{3x}\)-4\(\sqrt{3x}\)+27-3\(\sqrt{3x}\)=27-5\(\sqrt{3x}\)
b)3\(\sqrt{2x}\)-5\(\sqrt{8x}\)+7\(\sqrt{18x}\)+28
=3\(\sqrt{2x}\)-10\(\sqrt{2x}\)+21\(\sqrt{2x}\)+28
=14\(\sqrt{2x}\)+28=14(\(\sqrt{2x}\)+2)
a) \(2\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x}\)
\(=\left(2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}\right)+27\)
\(=-5\sqrt{3x}+27\)
Câu 1:
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
a) Thay x=16 vào B, ta được:
\(B=\dfrac{1}{\sqrt{16}-3}=\dfrac{1}{4-3}=1\)
Vậy: Khi x=16 thì B=1
b) Ta có: M=A-B
\(=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)
\(=\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+3\sqrt{x}-2\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)-2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\)
c) Để \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}-3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow x-4=x-2\sqrt{x}-3\)
\(\Leftrightarrow-2\sqrt{x}-3=-4\)
\(\Leftrightarrow-2\sqrt{x}=-1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)
hay \(x=\dfrac{1}{4}\)(thỏa ĐK)
Vậy: Để \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) thì \(x=\dfrac{1}{4}\)
Câu 2:
b) Gọi thời gian tổ 1 hoàn thành công việc khi làm một mình là x(giờ)
thời gian tổ 2 hoàn thành công việc khi làm một mình là y(giờ)
(Điều kiện: x>12; y>12)
Trong 1 giờ, tổ 1 làm được: \(\dfrac{1}{x}\)(công việc)
Trong 1 giờ, tổ 2 làm được: \(\dfrac{1}{y}\)(công việc)
Trong 1 giờ, hai tổ làm được: \(\dfrac{1}{12}\)(công việc)
Do đó, ta có phương trình: \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\)(1)
Vì khi tổ 1 làm trong 2 giờ, tổ 2 làm trong 7 giờ thì hai tổ hoàn thành được một nửa công việc nên ta có phương trình: \(\dfrac{2}{x}+\dfrac{7}{y}=\dfrac{1}{2}\)(2)
Từ (1) và (2) ta lập được hệ phương trình:
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{2}{x}+\dfrac{7}{y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=\dfrac{1}{6}\\\dfrac{2}{x}+\dfrac{7}{y}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-5}{y}=\dfrac{-1}{3}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=15\\\dfrac{1}{x}+\dfrac{1}{15}=\dfrac{1}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{60}\\y=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=60\\y=15\end{matrix}\right.\)(thỏa ĐK)
Vậy: Tổ 1 cần 60 giờ để hoàn thành công việc khi làm một mình
Tổ 2 cần 15 giờ để hoàn thành công việc khi làm một mình
a: \(B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
b: Để B=16 thì \(4\sqrt{x+1}=16\)
=>x+1=16
hay x=15
a: Khi x=16 thì \(A=\dfrac{6}{16-3\cdot4}=\dfrac{6}{4}=\dfrac{3}{2}\)
b: P=A:B
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{6}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
c: \(P-1=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}}=\dfrac{3}{\sqrt{x}}>0\)
=>P>1
\(a,B=4\sqrt{x=1}-3\sqrt{x+1}+2\)\(\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
\(b,\)đưa về \(\sqrt{x+1}=4\Rightarrow x=15\)
a, Với \(x\ge-1\)
\(\Rightarrow B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
b, Ta có B = 16 hay
\(4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\)bình phương 2 vế ta được
\(\Leftrightarrow x+1=16\Leftrightarrow x=15\)
a) B = 4√x+1 b) x = 15
) B=4 \sqrt{x+1}-3 \sqrt{x+1}+2 \sqrt{x+1}+\sqrt{x+1}=4 \sqrt{x+1}B=4x+1−3x+1+2x+1+x+1=4x+1.
b) Đưa về \sqrt{x+1}=4x+1=4. Suy ra x=15x=15.
a) \(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=\sqrt{16}.\sqrt{x+1}+\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}+\sqrt{x+1}=\sqrt{x+1}\left(4-3+2+1\right)=4\sqrt{x+1}\)
b) \(B=16\Leftrightarrow4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\Rightarrow x\in\left\{15\right\}\)
a) 4\(\sqrt{x+1}\) b)\(x=15\)
a) B= 4\(\sqrt{x-1}\)
b) x=15
a. B = \(4\sqrt{x+1}\)
b. Ta có
B = 16
⇔ \(\sqrt{x+1}\) = 4
⇒ x = 15
a) B=\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)=\(4\sqrt{x+1}\)
b)x=15
a) B = \(4\sqrt{x+1}\)
b) x=15
a/ B= 4\(\sqrt{x+1}\) - 3\(\sqrt{x+1}\) + 2\(\sqrt{x+1}\) + \(\sqrt{x+1}\) = 4\(\sqrt{x+1}\)
b/ x = 15
a. \(4\sqrt{x+1}\)
b. x=15
a)B=\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\) (với x\(\ge-1\))
B=\(\sqrt{4^2\cdot\left(x+1\right)}-\sqrt{3^2\cdot\left(x+1\right)}+\sqrt{2^2\cdot\left(x+1\right)}+\sqrt{x+1}\)
B=\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
B=\(4\sqrt{x+1}\)
Vậy B=\(4\sqrt{x+1}\) với x≥-1
b) Có B=4\(\sqrt{x+1}\) với x≥-1
Mà B = 16
⇔4\(\sqrt{x+1}\) =16
⇔\(\sqrt{x+1}=4\)
⇔x+1=16
⇔x=15(Thỏa mãn điều kiện )
Vậy x=15 thì B= 16
B=\(\sqrt{16x+16}\)- \(\sqrt{9x+9}\)+\(\sqrt{4x+4}\)+\(\sqrt{x+1}\) x\(\ge\)-1
a) B=4\(\sqrt{x+1}\)-3\(\sqrt{x+1}\)+2\(\sqrt{x+1}\)+\(\sqrt{x+1}\)
B= 4\(\sqrt{x+1}\)(TM)
Vậy với x\(\ge\)-1 thì B =4\(\sqrt{x+1}\)
b)Với x\(\ge-1\)thì B=16
4\(\sqrt{x+1}\)=16
\(\Leftrightarrow\)\(\sqrt{x+1}\)=4
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\)(TMDK)
Vậy với B=16 thì x=15
\(4\sqrt{x+1}\) b, x=15
a. B = \(\sqrt{16x+16}\) - \(\sqrt{9x+9}\) + \(\sqrt{4x+4}\) + \(\sqrt{x+1}\) với x \(\ge\)-1
B = 4\(\sqrt{x+1}\) - 3\(\sqrt{x+1}\) + 2\(\sqrt{x+1}\) + \(\sqrt{x+1}\)
B = 4\(\sqrt{x+1}\)
Vậy vs x\(\ge-1\) thì B = 4\(\sqrt{x+1}\)
b. vs x\(\ge-1\)
\(\Rightarrow\) B = 16
\(\Leftrightarrow4\sqrt{x+1}\) = 16
\(\Leftrightarrow\sqrt{x+1}\) = 4
\(\Leftrightarrow\) x+1 = 16
\(\Leftrightarrow\) x = 15 (tm)
vậy B = 16 thì x = 15
a) B=4\(\sqrt{x+1}\)-3\(\sqrt{x+1}\)+2\(\sqrt{x+1}\)+\(\sqrt{x+1}\)=4\(\sqrt{x+1}\)
b) B= 16
\(\Leftrightarrow\)4\(\sqrt{x+1}\)=16
\(\Leftrightarrow\)\(\sqrt{x+1}\)=4
\(\Leftrightarrow\)(\(\sqrt{x+1}\))\(^2\)=4\(^2\)
\(\Leftrightarrow\)x+1=16
\(\Leftrightarrow\)x=15
a) B=4 \sqrt{x+1}-3 \sqrt{x+1}+2 \sqrt{x+1}+\sqrt{x+1}=4 \sqrt{x+1}B=4x+1−3x+1+2x+1+x+1=4x+1.
b) Đưa về \sqrt{x+1}=4x+1=4. Suy ra x=15x=15.
bài 3
a) B=\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}với\ge-1\) b) B=16 theo đầu bài
B=\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\) => 16=\(4\sqrt{x+1}\)
B=\(\sqrt{x+1}\left(4-3+2+1\right)\) <=> 16/4=\(\sqrt{x+1}\)
B=\(4\sqrt{x+1}\) <=> 4=\(\sqrt{x+1}\)
<=> \(4^2=\left(\sqrt{x+1}\right)^2\)
<=> 16 = x+1
<=> 15=x
Vậy x=15
b, Để B = 16 thì
⇔ x + 1 = 16 ⇔ x = 15 (thỏa mãn x ≥ -1)
a)B=\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)
=\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
=\(4\sqrt{x+1}\)
b) Để B=16 thì \(4\sqrt{x+1}=16\) <=>\(\sqrt{x+1}=4\)
<=>\(x+1=16\)
<=>\(x=15\)(thỏa mãn đkxđ)
\(a)B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=\sqrt{4\left(x+1\right)}-\sqrt{3\left(x+1\right)}+\sqrt{2\left(x+1\right)}+\sqrt{x+1}=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=4\sqrt{x+1}\)
\(4\sqrt{x+1}=16\leftrightarrow\sqrt{x+1}=4\leftrightarrow x+1=16\leftrightarrow x=15\) (thỏa mãn x ≥ -1)
a,4\(\sqrt{x+1}\)
b,\(x=15\)