\(M=\left(x^2+y^2\right)^2+x^4+x^2y^2+y^2\)

\(M=1+x^2\left(x^2+y^2\right)+y^2\)

\(M=1+x^2+y^2\)

\(M=1+1=2\)

câu b bạn xem lại đề ạ chắc thiếu mất dấu cộng

\(4\left(3x^2+5x+2\right)=0\Leftrightarrow12x^2+20x+8=0\)

\(\Leftrightarrow12x^2+20x+1=-7\)

thanks bn nhaa, mk k cho bn rr

\(x^2+2x^2+3x^2+4x^2+5x^2=220\)

\(\Leftrightarrow x^2+2^2.x^2+3^2.x^2+4^2.x^2+5^2.x^2=220\)

\(\Leftrightarrow x^2\left(1+2^2+3^2+4^2+5^2\right)=220\)

\(\Leftrightarrow x^2\left(1+4+9+16+25\right)=220\)

\(\Leftrightarrow x^2.55=220\)

\(\Leftrightarrow x^2=\frac{220}{55}\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=2\left(x>0\right)\)

Vậy \(x=2\)

x^2+4x^2+9x^2+16x^2+25x^2=220

55x^2=220

x^2=220:55

x^2=4

x=2

Vậy x có giá trị bằng 2

a)x²-3x+2=0

=>x²-2x-x+2=0

=>x.(x-2)-(x-2)=0

=>(x-2)(x-1)=0

=>x-2=0 hoặc x-1=0

=>x=2 hoặc x=1

b)2x²-5x+3<0

=>2x²-2x-3x+3<0

=>2x.(x-1)-3.(x-1)<0

=>(x-1)(2x-3)<0

TH1: x-1 >0 và 2x-3<0

=>x>1 và x<3/2

=>1<x<3/2

TH2: x-1<0 và 2x-3>0

=>x<1 và x>3/2(vô lí)

Vậy 1<x<3/2

còn câu c bạn tự giải nha

a) Ta có: x² + 4x +5 = ( x² + 4x + 4 ) +1 =  (x+2)²  + 1  >= 1 >0 với mọi x

b) Ta có : 4x² - 4x +2 = ( 4x² - 4x +1 ) + 1 = (2x+1)²  > 0 với mọi x

c) Ta có : x² - 3x +4 = [x² - 2.(3/2)x + (9/4) ]+ (7/4) = ( x - 3/2 )² + 7/4 >0 với mọi x 

mấy câu sau lm tương tự: sử dụng hằng đẳng thức tách thành dạng một bình phương cộng vs 1 số 

a) x² + 4x + 5 = x² + 2 . 2x + 2² + 1 = (x + 2)² + 1\(\ge\)1 > 0

b) 4x² - 4x + 2 = (2x)² - 2 . 2x + 1 + 1 = (2x - 1)² + 1\(\ge\)1 > 0

c) x² - 3x + 4 = x² - 2 . 1,5x + 1,5² + 1,75 = (x - 1,5)² + 1,75 \(\ge\)1,75  > 0

d) x² - x + 1 = x² + 2 . 0,5x + 0,5² + 0,75 = (x + 0,5)² + 0,75\(\ge\)0,75  > 0

e) x² - 5x + 7 = x² - 2 . 2,5x + 2,5² + 0,75 = (x - 2,5)² + 0,75\(\ge\)0,75  > 0

a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

\(\Leftrightarrow x^2-x-2x+2=0\)

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết