x^2 - 4x + 4 = 5 ( x - 2 ) 

x^2 - 4x + 4 - 5 ( x - 2 ) = 0 

( x - 2 ) ^2 - 5 ( x - 2 ) = 0 

( x - 2 ) ( x - 2 - 5 ) = 0

( x - 2 ) ( x - 7 ) = 0 

x  - 2 = 0 hoặc x - 7 = 0 

x = 2 hoặc x = 7 

c) (x-2)^2=5(x-2)

=> x-2=5 hoặc x-2 =0

=> x=7 hoặc x=2

d) (2x-3)^2=(5-x)^2

=> 2x-3=5-x

=> x=8/3

4x^2 - 12x + 9 = ( 5 - x ) ^2 

( 2x - 3 ) ^2 = ( 5 - x ) ^2 

( 2x - 3 ) ^2 - ( 5 - x ) ^2 = 0 

( 2x - 3 - 5 + x ) ( 2x - 3 + 5 - x ) = 0 

( 3x - 8 ) ( x + 2 ) = 0 

3x - 8 = 0 hoặc x + 2 = 0 

x = 8/3 hoặc x = -2 

x^2+1>=1

=>(x^2+1)^2>=1

y^2+2>=2

=>(y^2+2)^4>=16

=>(x^2+1)^2+(y^2+2)^4>=17

=>(x^2+1)^2+(y^2+2)^4-2>=15

Dấu = xảy ra khi x=y=0

1) Ta có: \(\left(x^2-1\right)^2-x\left(x^2-1\right)-2x^2=0\)

\(\Leftrightarrow\left[\left(x^2-1\right)^2+x\left(x^2-1\right)\right]-\left[2x\left(x^2-1\right)+2x^2\right]=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-2x-1=0\\x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=2\\\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\pm\sqrt{2}\\x+\frac{1}{2}=\pm\frac{\sqrt{5}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\pm\sqrt{2}\\x=-\frac{1\pm\sqrt{5}}{2}\end{cases}}\)

2) Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)

\(\Leftrightarrow\left[\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)\right]+\left[2x\left(x^2+4x+8\right)+2x^2\right]=0\)

\(\Leftrightarrow\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)=0\)

\(\Leftrightarrow\left(x^2+6x+8\right)\left(x^2+5x+8\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)=0\)

Vì \(x^2+5x+8=\left(x^2+5x+\frac{25}{4}\right)+\frac{7}{4}=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}>0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)

Vậy x = -2 hoặc x = -4

P/s: Bạn Thùy Linh nếu PT chứa nghiệm vô tỉ thì với trình độ bình thường không dễ tìm được nghiệm đâu nhé

3) Ta có: \(4\left(x^2+x+1\right)^2+5x\left(x^2+x+1\right)+x^2=0\)

\(\Leftrightarrow\left[4\left(x^2+x+1\right)^2+4x\left(x^2+x+1\right)\right]+\left[x\left(x^2+x+1\right)+x^2\right]=0\)

\(\Leftrightarrow4\left(x^2+x+1\right)\left(x^2+2x+1\right)+x\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(4x^2+4x+4+x\right)\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(4x^2+5x+4\right)\left(x+1\right)^2=0\)

Xét PT \(4x^2+5x+4=0\) ta có:

\(\Delta_x=5^2-4\cdot4\cdot4=-39< 0\)

\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x=-1\)

Vậy x = -1

ta có ;

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

\(=\left(2x\right)^2-4x^2=0\)

\(\Rightarrow P=x^2-2xy+y^2+x^2+2xy+y^2-2\left(x^2-y^2\right)-4x^2\)\(4x^2\)

\(\Rightarrow P=2x^2+2y^2-2x^2-2y^2-4x^2\)

\(\Rightarrow P=-4x^2\)

TA CÓ:

\(a,\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\Leftrightarrow\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\left(x-3\right)\left(4x-1-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-x-3\right)=0\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

\(b,\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\Leftrightarrow\left(x+3\right)\left(x-5+3x-4\right)=0\)

\(\left(x-3\right)\left(4x-9\right)=0\orbr{\begin{cases}x=3\\x=\frac{9}{4}\end{cases}}\)

\(c,\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\Leftrightarrow\left(1-x\right)\left(5x+3\right)=\left(7-3x\right)\left(1-x\right)\)

\(\left(1-x\right)\left(5x+3-7+3x\right)=0\Leftrightarrow\left(1-x\right)\left(8x-4\right)=0\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

ĐKXĐ: \(2x-y-1\ge0;x+2y\ge0\)

Đặt \(\sqrt{2x-y-1}=a;\sqrt{x+2y}=b\left(a,b\ge0\right)\). Khi đó ta có:

\(\left(2b^2-1\right)a=\left(2a^2-1\right)b\Leftrightarrow\left(a-b\right)\left(2ab+1\right)=0\)

\(\Leftrightarrow a=b\) hoặc \(2ab+1=0\)(loại vì \(a,b\ge0\))

Suy ra: \(\sqrt{2x-y-1}=\sqrt{x+2y}\Leftrightarrow x=3y+1\)

Pt đầu tiên trở thành: \(\left(3y+1\right)^2-5y^2-8y=3\)

\(\Leftrightarrow\left(y-1\right)\left(2y+1\right)=0\Leftrightarrow\orbr{\begin{cases}y=1\\y=-\frac{1}{2}\end{cases}}\)

+) Với  \(y=1\Rightarrow x=4\Rightarrow\left(x;y\right)=\left(4;1\right)\)(tm)

+) Với  \(y=-\frac{1}{2}\Rightarrow x=-\frac{1}{2}\Rightarrow\left(x;y\right)=\left(-\frac{1}{2};-\frac{1}{2}\right)\) (loại)

Vậy hpt có nghiệm duy nhất \(\left(x;y\right)=\left(4;1\right).\)