bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

a) \(x^3-4x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}}\)

b) \(5x\left(3x-2\right)=4-9x^2\)

\(5x\left(3x-2\right)-\left(4-9x^2\right)=0\)

\(5x\left(3x-2\right)-\left(2-3x\right)\left(2+3x\right)=0\)

\(5x\left(3x-2\right)+\left(3x-2\right)\left(2+3x\right)=0\)

\(\left(3x-2\right)\left(5x+3x+2\right)=0\)

\(\left(3x-2\right)\left(8x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\8x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-1}{4}\end{cases}}}\)

c) \(x^2+7x=8\)

\(x^2+7x-8=0\)

\(x^2+8x-x-8=0\)

\(x\left(x+8\right)-\left(x+8\right)=0\)

\(\left(x+8\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+8=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-8\\x=1\end{cases}}}\)

d) \(2x^2+4y^2+10x+4xy=-25\)

\(x^2+x^2+4y^2+10x+4xy+25=0\)

\(\left(4y^2+4xy+x^2\right)+\left(x^2+10x+25\right)=0\)

\(\left(2y+x\right)^2+\left(x+5\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}2y+x=0\\x+5=0\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{5}{2}\\x=-5\end{cases}}}\)

\(\text{Tìm x:}\)

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

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

\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)

\(b.3x\left(x-2\right)+10-5x=0\)

\(3x^2-6x+10-5x=0\)

\(3x^2-11x+10=0\)

\(3x^2-11x=-10\)(bn xem lại đề nhé)

\(c.x^3-5x^2+x-5=0\)

\(x^3-5x^2+x=5\)

\(d.x^4-2x^3+10x^2-20x=0\)

bài 1:phân tích thành phân tử

  a> x^2-6x-y^2+9

= (x-3)^2 -y^2

= (x-3 -y) (x-3+y)

b>x^2-xy-8x+8y

= x(x-y) - 8(x-y)

= (x-8) (x-y)

c>25-4x^2-4xy-y^2

= 5^2 - (2x + y)^2 

= (5 - 2x -y) (5 +2x+y) 

d>xy-xz-y+z

= x(y-z) - (y-z)

= (x-1) (y-z)

e>x^2-xz-yz+2xy+y^2

= (x+y)^2 - z(x+y)

= (x+y-z) (x+y)

g>x^2-4xy+4y^2-z^2-4zt-4t^2

= (x-2y)^2 - (z + 2t)^2 

= (x-2y -x-2t) (x-2y + z +2t)

bài 2:tìm X bt 

a>x.(x-1)-3x+3x=0

x (x-1) =0

\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy x=0 và x=1

b>3x.(x-2)+10-5x=0

3x(x-2) - 5 (x-2)=0

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

\(\Rightarrow\hept{\begin{cases}3x-5=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}3x=5\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)

c>x^3-5x^2+x-5=0

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

(x^2 +1)(x-5) =0

\(\Rightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x^2=-1\\x=5\end{cases}\Rightarrow}\hept{\begin{cases}x\in\varphi\\x=5\end{cases}}}\)

Vậy x=5

d>x^4-2x^3+10x^2-20x=0

x^3 (x-2) + 10x(x-2) =0 

(x^3 + 10x) (x-2) =0

x(x^2 + 10) (x-2) =0

\(\Rightarrow\hept{\begin{cases}x=0\\x^2+10=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-10\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x\in\varphi\\x=2\end{cases}}}}\)

Vậy x=0 và x=2

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

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

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

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

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

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

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

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

Bài 2:

\(A=x^2+4y^2-2x+10-4xy-4y\)

\(=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

Thay x + 2y = 5 vào biểu thức A ta được: \(A=5^2-2.5+10=25\)

\(B=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)\left(y-1\right)+y^2-2y+1\)

\(=x^2+4xy+4y^2-2xy+2x-4y^2+4y+y^2-2y+1\)

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

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

Thay x + y = 5 vào biểu thức B ta được: \(B=5^2+2.5+1=25+10+1=36\)

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

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

Thay x + y = 2 vào C ta được: \(C=\left(x-2-y\right)\left(2-2\right)-4=0-4=-4\)

\(D=x^2+y^2+2xy-4x-4y-3\)

\(=\left(x+y\right)^2-4\left(x+y\right)-3\) Thay x + y = 4 vào D ta được:

\(D=4^2-4.4-3=16-16-3=-3\)

Bài 3:

a) \(N=-9x^2+12x-5=-\left(9x^2-12x+4\right)-1\)

\(=-\left(3x-2\right)^2-1\)

Do \(\left(3x-2\right)^2\ge0\) nên \(-\left(3x-2\right)^2-1< 0\)

Vậy N < 0

b) ghi đề cẩn thận lại đi, mk k hiểu

a/ \(4x^2+2y^2-4xy+4x-2y+5=0\)

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

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

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

Với mọi x, y ta có :

\(\left(2x-y+1\right)^2\ge0\Leftrightarrow\left(2x-y+1\right)^2+4>0\)

\(\Leftrightarrow pt\) vô nghiệm