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Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
a: TH1: m=1
Pt sẽ là -(2*1-1)x+1+1=0
=>-x+2=0
=>x=2(loại)
TH2: m<>1
\(\text{Δ}=\left(2m-1\right)^2-4\left(m-1\right)\left(m+1\right)\)
\(=4m^2-4m+1-4m^2+4=-4m+5\)
Để phương trình có hai nghiệm phân biệt thì -4m+5>0
=>m<5/4
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}x_1-3x_2=0\\x_1+x_2=\dfrac{2m-1}{m-1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x_2=\dfrac{-2m+1}{m-1}\\x_1=3x_2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{2m-1}{4\left(m-1\right)}\\x_1=\dfrac{6m-3}{4m-4}\end{matrix}\right.\)
x1x2=m+1/m-1
=>\(\dfrac{\left(2m-1\right)\left(6m-3\right)}{16\left(m-1\right)^2}=\dfrac{m+1}{m-1}\)
=>\(\dfrac{\left(2m-1\right)\left(6m-3\right)}{16\left(m-1\right)^2}=\dfrac{16\left(m-1\right)\left(m+1\right)}{16\left(m-1\right)^2}\)
=>\(16m^2-16=12m^2-12m+3\)
=>4m^2+12m-19=0
hay \(x=\dfrac{-3\pm2\sqrt{7}}{2}\)
c: \(\text{Δ}=\left(2m-2\right)^2-12\left(3m-5\right)\)
\(=4m^2-8m+4-36m+60=4m^2-44m+64\)
Để phương trình có hai nghiệm phân biệt thì m^2-11m+16>0
=>\(\left\{{}\begin{matrix}x< \dfrac{11-\sqrt{57}}{2}\\x>\dfrac{11+\sqrt{57}}{2}\end{matrix}\right.\)
Theo đề, ta có hệ:
x1-x2=0 và x1+x2=2m-2/3
=>2x1=(2m-2)/3 và x1=x2
=>x1=x2=m-1/3
x1*x2=3m-5/3
=>\(\dfrac{m^2-2m+1}{9}=\dfrac{3m-5}{3}\)
=>m^2-2m+1=9m-15
=>m^2-11m+16=0
hay \(m\in\varnothing\)
1.
Đặt \(x^2-2x+m=t\), phương trình trở thành \(t^2-2t+m=x\)
Ta có hệ \(\left\{{}\begin{matrix}x^2-2x+m=t\\t^2-2t+m=x\end{matrix}\right.\)
\(\Rightarrow\left(x-t\right)\left(x+t-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=t\\x=1-t\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=x^2-2x+m\\x=1-x^2+2x-m\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=-x^2+3x\\m=-x^2+x+1\end{matrix}\right.\)
Phương trình hoành độ giao điểm của \(y=-x^2+x+1\) và \(y=-x^2+3x\):
\(-x^2+x+1=-x^2+3x\)
\(\Leftrightarrow x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{4}\)
Đồ thị hàm số \(y=-x^2+3x\) và \(y=-x^2+x+1\):
Dựa vào đồ thị, yêu cầu bài toán thỏa mãn khi \(m< \dfrac{5}{4}\)
Mà \(m\in\left[-10;10\right]\Rightarrow m\in[-10;\dfrac{5}{4})\)
Có cách nào lm bài này bằng cách lập bảng biến thiên k ạ
`#3107.101107`
a,
\(\text{A = }\left\{x\in R\text{ | }\left(2x-x^2\right)\left(3x-2\right)=0\right\}\)
`<=> (2x - x^2)(3x - 2) = 0`
`<=>`\(\left[{}\begin{matrix}2x-x^2=0\\3x-2=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x\left(2-x\right)=0\\3x=2\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2-x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy, `A = {0; 2; 2/3}`
b,
\(\text{B = }\left\{x\in R\text{ | }2x^3-3x^2-5x=0\right\}\)
`<=> 2x^3 - 3x^2 - 5x = 0`
`<=> x(2x^2 - 3x - 5) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-3x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-2x+5x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x^2-2x\right)+\left(5x-5\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x\left(x-1\right)+5\left(x-1\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x+5\right)\left(x-1\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x+5=0\\x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)
Vậy, `B = {-5/2; 0; 1}.`
c,
\(\text{C = }\left\{x\in Z\text{ | }2x^2-75x-77=0\right\}\)
`<=> 2x^2 - 75x - 77 = 0`
`<=> 2x^2 - 2x + 77x - 77 = 0`
`<=> (2x^2 - 2x) + (77x - 77) = 0`
`<=> 2x(x - 1) + 77(x - 1) = 0`
`<=> (2x + 77)(x - 1) = 0`
`<=>`\(\left[{}\begin{matrix}2x+77=0\\x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=-77\\x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=-\dfrac{77}{2}\\x=1\end{matrix}\right.\)
Vậy, `C = {-77/2; 1}`
d,
\(\text{D = }\left\{x\in R\text{ | }\left(x^2-x-2\right)\left(x^2-9\right)=0\right\}\)
`<=> (x^2 - x - 2)(x^2 - 9) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-x-2=0\\x^2-9=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2+x-2x-2=0\\x^2=9\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}\left(x^2+x\right)-\left(2x+2\right)=0\\x^2=\left(\pm3\right)^2\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x\left(x+1\right)-2\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x-2=0\\x+1=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=2\\x=-1\\x=\pm3\end{matrix}\right.\)
Vậy, `D = {-1; -3; 2; 3}.`
Câu 2:
\(\left(A\cup B\right)\cap C=A\cap C=[1;+\infty)\cap\left(0;4\right)=[1;4)\)
Tập này có 3 phần tử nguyên
1) \(x^2-2mx+m-2=0\) (1)
pt (1) có \(\Delta'=\left(-m\right)^2-\left(m-2\right)=m^2-m+2=\left(m-\frac{1}{2}\right)^2+\frac{7}{4}>0\left(\forall m\right)\)
=> pt luôn có 2 nghiệm phân biệt x1, x2
Vi-et: \(\hept{\begin{cases}x_1+x_2=2m\\x_1x_2=m-2\end{cases}}\)\(\Rightarrow\)\(M=\frac{2x_1x_2-\left(x_1+x_2\right)}{x_1^2+x_2^2-6x_1x_2}=\frac{2x_1x_2-\left(x_1+x_2\right)}{\left(x_1+x_2\right)^2-8x_1x_2}=\frac{2m-4-2m}{\left(2m\right)^2-8m-16}\)
\(=\frac{-4}{4m^2-8m-16}=\frac{-4}{4\left(m-1\right)^2-20}\ge\frac{-4}{-20}=\frac{1}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(m=1\)
xin 1slot sáng giải
a: ĐKXĐ: \(x^2-6x+6\ge0\)
=>\(x^2-6x+9-3\ge0\)
=>\(\left(x-3\right)^2-3\ge0\)
=>\(\left(x-3\right)^2\ge3\)
=>\(\left[\begin{array}{l}x-3\ge\sqrt3\\ x-3\le-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\sqrt3+3\\ x\le-\sqrt3+3\end{array}\right.\)
Ta có: \(x^2-6x+9=4\sqrt{x^2-6x+6}\)
=>\(x^2-6x+6-4\cdot\sqrt{x^2-6x+6}+3=0\)
=>\(\left(\sqrt{x^2-6x+6}-3\right)\left(\sqrt{x^2-6x+6}-1\right)=0\)
TH1: \(\sqrt{x^2-6x+6}-3=0\)
=>\(\sqrt{x^2-6x+6}=3\)
=>\(x^2-6x+6=9\)
=>\(x^2-6x-3=0\)
=>\(x^2-6x+9-12=0\)
=>\(\left(x-3\right)^2=12\)
=>\(\left[\begin{array}{l}x-3=2\sqrt3\\ x-3=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3+3\left(nhận\right)\\ x=3-2\sqrt3\left(nhận\right)\end{array}\right.\)
TH2: \(\sqrt{x^2-6x+6}-1=0\)
=>\(x^2-6x+6=1\)
=>\(x^2-6x+5=0\)
=>(x-1)(x-5)=0
=>\(\left[\begin{array}{l}x=1\left(nhận\right)\\ x=5\left(nhận\right)\end{array}\right.\)
b: ĐKXĐ: x∈R
\(x^2-x+8-4\sqrt{x^2-x+4}=0\)
=>\(x^2-x+4-4\cdot\sqrt{x^2-x+4}+4=0\)
=>\(\left(\sqrt{x^2-x+4}-2\right)^2=0\)
=>\(\sqrt{x^2-x+4}-2=0\)
=>\(\sqrt{x^2-x+4}=2\)
=>\(x^2-x+4=4\)
=>\(x^2-x=0\)
=>x(x-1)=0
=>x=0 hoặc x=1
c: \(x^2+\sqrt{4x^2-12x+44}=3x+4\)
=>\(x^2-3x-4+2\sqrt{x^2-3x+11}=0\)
=>\(x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)
=>\(\left(\sqrt{x^2-3x+11}+5\right)\left(\sqrt{x^2-3x+11}-3\right)=0\)
=>\(\sqrt{x^2-3x+11}-3=0\)
=>\(\sqrt{x^2-3x+11}=3\)
=>\(x^2-3x+11=9\)
=>\(x^2-3x+2=0\)
=>(x-1)(x-2)=0
=>x=1(nhận) hoặc x=2(nhận)