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a)\(-\frac{2}{5}+\frac{2}{3}x+\frac{1}{6}x=-\frac{4}{5}\Leftrightarrow\frac{5}{6}x=-\frac{2}{5}\Leftrightarrow x=-\frac{12}{25}\)
Vậy nghiệm là x = -12/25
b)\(\frac{3}{2}x-\frac{2}{5}-\frac{2}{3}x=-\frac{4}{15}\Leftrightarrow\frac{5}{6}x=\frac{2}{15}\Leftrightarrow x=\frac{4}{25}\)
Vậy nghiệm là x = 4/25
c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)\(\Leftrightarrow x=-1\)
Vậy nghiệm là x = -1
bài 1 :
b) (x-1/2 )2 = 0
<=> x - 1/2 = 0
<=> x = 0+ 1/2
<=> x = 1/2
c) ( x - 2 ) 2 = 1
<=> x -2 = 1
<=> x = 1 +2 = 3
d) ( 2x -1 )3 = -8
<=> ( 2x - 1) 3 = ( -2 ) 3
<=> 2x - 1 = -2
<=> 2x = -2+1 = -1
<=> x = -1/2
Bài 2 :
c) 32x-1=243
<=> 32x-1= 35
<=> 2x-1 = 5
<=> 2x = 6
<=> x = 6:2 = 3
Mk chỉ giải đc như vậy thôi
bạn thông cảm nhé !
Bạn ghi ra nhiều vậy người khác nhìn rối mắt không trả lời được đâu ghi từng bài ra thôi
Mình chỉ làm được vài bài thôi, kiến thức có hạn :>
Bài 1:
Câu a và c đúng
Bài 2:
a) |x| = 2,5
=>x = 2,5 hoặc
x = -2,5
b) |x| = 0,56
=>x = 0,56
x = - 0,56
c) |x| = 0
=. x = 0
d)t/tự
e) |x - 1| = 5
=>x - 1 = 5
x - 1 = -5
f) |x - 1,5| = 2
=>x - 1,5 = 2
x - 1,5 = -2
=>x = 2 + 1,5
x = -2 + 1,5
=>x = 3,5
x = - 0,5
các câu sau cx t/tự thôi
Bài 3: Ko hỉu :)
Bài 4: Kiến thức có hạn :)
Câu a: - \(\frac34x\) - 3 = \(\frac14x\)
- \(\frac34x\) - \(\frac14x\) = 3
-\(x=3\)
\(x=3:\left(-1\right)\)
\(x\) = -3
Vậy \(x=-3\)
Câu b:
2,8 : 0,5 = |\(x-1\)| : 1,5
5,6 = |\(x-1\)| : 1,5
|\(x-1\)| = 5,6 x 1,5
|\(x-1\)| = 8,4
\(\left[\begin{array}{l}x-1=-8,4\\ x-1=8,4\end{array}\right.\)
\(\left[\begin{array}{l}x=-8,4+1\\ x=8,4+1\end{array}\right.\)
\(\left[\begin{array}{l}x=-7,4\\ x=9,4\end{array}\right.\)
Vậy \(x\in\) {-7;4 ;9,4}
a/ \(x+\dfrac{3}{5}=\dfrac{4}{7}\)
\(x=\dfrac{4}{7}-\dfrac{3}{5}\)
\(x=-\dfrac{1}{35}\)
Vậy ....
b/ \(x-\dfrac{5}{6}=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+\dfrac{5}{6}\)
\(x=1\)
Vậy ....
c/\(-\dfrac{5}{7}-x=\dfrac{-9}{10}\)
\(x=\dfrac{-5}{7}-\dfrac{-9}{10}\)
\(x=\dfrac{13}{70}\)
Vậy .....
d/ \(\dfrac{5}{7}-x=10\)
\(x=\dfrac{5}{7}-10\)
\(x=\dfrac{-65}{7}\)
Vậy ...
e/ \(x:\left(\dfrac{1}{9}-\dfrac{2}{5}\right)=\dfrac{-1}{2}\)
\(x:\dfrac{-13}{45}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}.\dfrac{-13}{45}\)
\(x=\dfrac{13}{90}\)
Vậy ....
f/ \(\left(\dfrac{-3}{5}+1,25\right)x=\dfrac{1}{3}\)
\(0,65.x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:0,65\)
\(x=\dfrac{20}{39}\)
Vậy ....
g/ \(\dfrac{1}{3}x+\left(\dfrac{2}{3}-\dfrac{4}{9}\right)=\dfrac{-3}{4}\)
\(\dfrac{1}{3}x+\dfrac{2}{9}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{-35}{36}\)
\(\Leftrightarrow x=\dfrac{-35}{12}\)
Vậy ...
a: \(\Rightarrow\left(2x-4\right)^{x+1}\left[\left(2x-4\right)^4-1\right]=0\)
=>(2x-4)(2x-3)(2x-5)=0
hay \(x\in\left\{2;\dfrac{3}{2};\dfrac{5}{2}\right\}\)
b: \(\Leftrightarrow\left(x-3\right)^{x+4}\left(x-3-1\right)=0\)
=>(x-3)x+4(x-4)=0
=>x=3 hoặc x=4
c: \(\Leftrightarrow\left[{}\begin{matrix}x-1>2\\x-1< -2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\)
d: =>-5<=2x+3<=5
=>-8<=2x<=2
=>-4<=x<=1
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
1) \(\left|x\right|=7\)
=> \(\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
Vậy \(x\in\left\{7;-7\right\}.\)
2) \(\left|x\right|=0\)
=> \(x=0\)
Vậy \(x\in\left\{0\right\}.\)
5) \(\left|x\right|-1=\frac{2}{5}\)
=> \(\left|x\right|=\frac{2}{5}+1\)
=> \(\left|x\right|=\frac{7}{5}\)
=> \(\left[{}\begin{matrix}x=\frac{7}{5}\\x=-\frac{7}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{7}{5};-\frac{7}{5}\right\}.\)
8) \(\left|x-17\right|=23\)
=> \(\left[{}\begin{matrix}x-17=23\\x-17=-23\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=23+17\\x=\left(-23\right)+17\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=40\\x=-6\end{matrix}\right.\)
Vậy \(x\in\left\{40;-6\right\}.\)
Mình chỉ làm thế thôi nhé, bạn đăng hơi nhiều mà với cả mấy câu này dễ mà bạn.
Chúc bạn học tốt!
1) |x|=7
=> [x=7x=−7 =>[x=7x=−7
Vậy x∈{7;−7}.x∈{7;−7}.
2) |x|=0
=> x=0x=0
Vậy x∈{0}.x∈{0}.
5) |x|−1=25
=> |x|=25+1 =>|x|=25+1
=> |x|=75|x|=75
=> [x=75x=−75[x=75x=−75
Vậy x∈{75;−75}.x∈{75;−75}.
8) |x−17|=23
=> [x−17=23x−17=−23[x−17=23x−17=−23 => [x=23+17x=(−23)+17[x=23+17x=(−23)+17
=> [x=40x=−6[x=40x=−6
Vậy x∈{40;−6}.
mình làm tới đây thôi dài quá:)
tick cho mình nha
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
a)
=> \(x+2=69\)
=> \(x=67\)
b)
=> \(2^{x-5}=2^{30}\)
=> \(x-5=30\)
=> \(x=35\)
c)
=> \(3^x\left(3^2+1\right)=810\)
=> \(3^x.10=810\)
=> \(3^x=81\)
=> \(x=4\)
d)
=> \(5^x\left(5-1\right)=500\)
=> \(5^x.4=500\)
=> \(5^x=125\)
=> \(x=3\)
a) 3x + 2 = 369
=> x + 2 = 69
=> x = 67
b) 2x - 5 = 810
=> 2x - 5 = (23)10
=> 2x - 5 = 230
=> x - 5 = 30
=> x =35
c) 3x + 2 + 3x = 810
=> 3x(32 + 1) = 810
=> 3x.10 = 810
=> 3x = 81
=> 3x = 34
=> x = 4
d) 5x + 1 - 5x = 500
=> 5x(5 - 1) = 500
=> 5x.4 = 500
=> 5x = 125
=> 5x = 53
=> x = 3
a: Ta có: \(\frac52x-\frac13=\frac43\)
=>\(\frac52x=\frac43+\frac13=\frac53\)
=>\(x=\frac53:\frac52=\frac23\)
b: \(\frac25+\frac35:x=\frac13\)
=>\(\frac35:x=\frac13-\frac25=\frac{5}{15}-\frac{6}{15}=\frac{-1}{15}\)
=>\(x=\frac35:\frac{-1}{15}=\frac35\cdot\left(-15\right)=-9\)
c: \(2\left(x+\frac12\right)^2=32\)
=>\(\left(x+\frac12\right)^2=\frac{32}{2}=16\)
=>\(\left[\begin{array}{l}x+\frac12=4\\ x+\frac12=-4\end{array}\right.=>\left[\begin{array}{l}x=4-\frac12=\frac72\\ x=-4-\frac12=-\frac92\end{array}\right.\)
d: \(3^{x+2}+3^{x}=810\)
=>\(3^{x}\cdot9+3^{x}=810\)
=>\(3^{x}\left(9+1\right)=810\)
=>\(3^{x}=\frac{810}{10}=81=3^4\)
=>x=4
a/ 5/2x -1/3 =4/3
5/2.x=4/3+1/3
5/2.x=5/3
x=5/3:5/2
x=5/3.2/5
x=10/15=2/3
a)
\(\frac{5}{2} x - \frac{1}{3} = \frac{4}{3} \textrm{ }\textrm{ } \Longrightarrow \textrm{ }\textrm{ } \frac{5}{2} x = \frac{5}{3} \textrm{ }\textrm{ } \Longrightarrow \textrm{ }\textrm{ } x = \frac{2}{3}\)
b)
\(\frac{2}{5} + \frac{3 / 5}{x} = \frac{1}{3} \textrm{ }\textrm{ } \Longrightarrow \textrm{ }\textrm{ } \frac{3 / 5}{x} = \frac{1}{3} - \frac{2}{5} = - \frac{1}{15} \textrm{ }\textrm{ } \Longrightarrow \textrm{ }\textrm{ } \frac{3}{5 x} = - \frac{1}{15} \textrm{ }\textrm{ } \Longrightarrow \textrm{ }\textrm{ } x = - 9\)
c)
\(2\left(\left(\right.x+\frac{1}{2}\left.\right)\right)^2=32\textrm{ }\Longrightarrow\textrm{ }\left(\right.x+1/2\left.\right)^2=16\textrm{ }\Longrightarrow\textrm{ }x+1/2=\pm4\textrm{ }\Longrightarrow\textrm{ }x=7.5\text{ ho}ặcx=-4.5\)
d)
\(3^{x + 2} + 3^{x} = 810 \textrm{ }\textrm{ } \Longrightarrow \textrm{ }\textrm{ } 9 \cdot 3^{x} + 3^{x} = 10 \cdot 3^{x} = 810 \textrm{ }\textrm{ } \Longrightarrow \textrm{ }\textrm{ } 3^{x} = 81 \textrm{ }\textrm{ } \Longrightarrow \textrm{ }\textrm{ } x = 4\)