\(\frac{-6xy\left(x+y\right)^2}{8x^3y\left(x+y\right)}=\frac{-3\left(x+y\right)}{4x^2}\)

áp dụng bài hc sẽ ra

mờ quá

a, \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

b, \(27x^3+27x^2+9x+1=0\Leftrightarrow27x^3+1+27x^2+9x=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2-3x+1\right)+9x\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2+2>0\right)=0\Leftrightarrow x=-\frac{1}{3}\)

c, \(9x^2\left(x+1\right)-4\left(x+1\right)=0\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{2}{3};x=\frac{2}{3};x=-1\)

d, \(\left(x+1\right)^3-25\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2-25\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+6\right)=0\Leftrightarrow x=-1;x=-6;x=4\)

a) \(\left(x+2\right)^2=x+2\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)

b) \(x^3+4x=0\)

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

\(\Leftrightarrow x=0\)(vì \(x^2+4>0\))

c) \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(2x+3\right)-\left(2x-3\right)\right]=0\)

\(\Leftrightarrow6\left(x-1\right)=0\)

\(\Leftrightarrow x=1\).

\(a,\left(x+2\right)^2=x+2\)

\(TH1:x=-2\)

\(0^2=0\)( luôn đúng)

\(TH2:x\ne-2\)

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

\(x+2=1\)

\(x=-1\)

\(b,x^3+4x=0\)

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

\(\orbr{\begin{cases}x=0\\x^2+4=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x^2=-4\left(KTM\right)\end{cases}}}\)

\(c,\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x+3-2x+3\right)=0\)

\(6\left(x-1\right)=0\)

\(x=1\left(TM\right)\)

D = x² + 4xy + 4y² - z² + 2xt - t² 

 = (x + 2y)² - (z - t)² 

= (x + 2y - z + t)(x + 2y + z - t) 

Thay x = 10 ; y = 40 ; z = 30 ; t = 20 vào D 

\(\Rightarrow D=\left(10+40.2-30+20\right)\left(10+40.2+30-20\right)=80.100=8000\)

D = x\(^2\) + 4xy + 4y \(^2\) - z \(^2\) + 2zt - t \(^2\)

D = (x + 2y)\(^2\) - z\(^2\)+ z\(^2\) + 2zt + t\(^2\) - t\(^2\) 

D = (10 + 80)\(^2\) - 30\(^2\) + (z + t)\(^2\) - 20\(^2\) 

D = 90\(^2\) - 900 - 900 + (30 + 20)\(^2\) - 400

D = 8100 - 900 + 2500 - 400 

D =8600

HT

Thực hiện phép chia đa thức: 

\(x^4-3x^3-7x^2+ax+b=\left(x^2-2x+6\right)\left(x^2-x-15\right)+\left(a-24\right)x+\left(b+90\right)\)

Khi đó: \(\hept{\begin{cases}a-24=3\\b+90=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=27\\b=-88\end{cases}}\)

\(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)+8\)

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

\(=\left(x^2+6x+4\right)^2-4^2+8\)

\(=\left(x^2+6x+4\right)^2-8\ge-8\)

Dấu \(=\)khi \(x^2+6x+4=0\Leftrightarrow x=-3\pm\sqrt{5}\).

\(B=5+\left(1-x\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=5-\left[\left(x-1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]\)

\(=5-\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=5-\left(x^2+5x\right)^2+6^2\)

\(=41-\left(x^2+5x\right)^2\le41\)

Dấu \(=\)khi \(x^2+5x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

\(C=\left(x+3\right)^4+\left(x-7\right)^4=\left[\left(x-2\right)+5\right]^4+\left[\left(x-2\right)-5\right]^4\)

\(=2\left(x-2\right)^4+300\left(x-2\right)^2+1250\ge1250\)

Dấu \(=\)khi \(x-2=0\Leftrightarrow x=2\).