\(2\left(x-1\right)-5\left(x+2\right)=-10\\ \Rightarrow2x-2-5x-10=-10\\ \Rightarrow-3x=2\\ \Rightarrow x=\dfrac{-2}{3}\)

Vậy \(x=\dfrac{-2}{3}\)

x:2=y:3->x/2=y/3

Áp dụng tính chất dãy tỉ số bằng nhau,ta có:

x/2=y/3=x+y/2+3=10/5=2

từ: x/2=2->x=2.2=4

y/3=2->y=2.3=6

vậy...

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{10}{5}=2\) (theo tính chất củadãy tỉ số bằng nhau)

\(\Rightarrow x=4,y=6\)

Ta có: \(x:2=y:3\)

nên \(\dfrac{x}{2}=\dfrac{y}{3}\)

mà x+y=10

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{10}{5}=2\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=2\\\dfrac{y}{3}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)

Vậy: x=4; y=6

\(\left(2x-1\right)^2=81\\ \Rightarrow\left(2x-1\right)=\sqrt{81}=9\\ \Rightarrow2x=9+1=10\\ \Rightarrow x=\dfrac{10}{2}=5\)

\(2^3.5-\left(x+3^2\right)=10\\ \Rightarrow8.5-\left(x+9\right)=10\\ \Rightarrow x+9=8.5-10=40-10=30\\ \Rightarrow x=30-9=21\)

help me

a)\(x^4+6x^3+11x^2+6x+1\)

\(=x^4+9x^2+1+6x^3+6x+2x^2\)

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

\(x^4+5x^3-12x^2+5x+1\)

\(=\left(x^4-2x^3+x^2\right)+\left(7x^3-14x^2+7x\right)+\left(x^2-2x+1\right)\)

\(=x^2\left(x^2-2x+1\right)+7x\left(x^2-2x+1\right)+\left(x^2-2x+1\right)\)

\(=\left(x^2+7x+1\right)\left(x^2-2x+1\right)\)

\(=\left(x^2+7x+1\right)\left(x-1\right)^2\)

\(4^x=32^{10}\)

\(\Rightarrow4^x=4^{80}\)

\(\Rightarrow x=80\)

\(4^x=32^{10}\)

\(\Rightarrow2^{2x}=2^{50}\)

\(\Rightarrow2x=50\Rightarrow x=25\)

help me

dễ mà bạn xin 20 phút làm ra giấy nhé :)) 

a) \(\left(x^4+6x^3+9x^2\right)+2x^2+6x+1\)

      \(\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)

        \(\left(x^2+3x+1\right)^2\)

b) \(x^4+x^3+x^2+x+1\)

câu b,  chúa sẽ c/m x ko tồn tại , và nó là 1 đa thức bất khả Q . trong R 

vì lớp 8 chưa học đến số phức  

     \(x^4+x^3=-x^2-x-1\)

 \(x^4+x^3+\frac{1}{4}x^2=\left(\frac{1}{4}x^2-x^2\right)-x-1\)

\(\left(x^2+\frac{1}{2}x\right)^2=-\frac{3}{4}x^2-x-1\)

\(4\left(x^2+\frac{1}{2}x\right)^2=-3x^2-4x-4\)

\(\Delta`=\left(-2\right)^2-\left(-4\right).\left(-3\right)=4-12< 0\)

          denta < 0 x vô nghiệm

vậy đa thức trên ko thể phân tích và nó là 1 đa thức bất khả Q

c) ,   

\(\left(6x^4-12x^3\right)+\left(17x^3-34x^2\right)-\left(4x^2-8x\right)-\left(3x-6\right)\)

\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)

     \(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)\)

      \(\left(x-2\right)\left\{\left(6x^3+18x^2\right)-\left(x^2+3x\right)-\left(x+3\right)\right\}\)

      \(\left(x-2\right)\left\{6x^2\left(x+3\right)-x\left(x+3\right)-\left(x+3\right)\right\}\)

\(\left(x-2\right)\left(x+3\right)\left(6x^2-x-1\right)\)

 \(\left(x-2\right)\left(x+3\right)\left\{\left(6x^2+\frac{6}{3}x\right)-\left(\frac{9}{3}x+\frac{9}{9}\right)\right\}\)

\(\left(x-2\right)\left(x+3\right)\left\{6x\left(x+\frac{1}{3}\right)-\frac{9}{3}\left(x+\frac{1}{3}\right)\right\}\)

\(\left(X-2\right)\left(X+3\right)\left(X+\frac{1}{3}\right)\left(6x-1\right)\)

d)

\(\left(x^4-x^3\right)+\left(6x^3-6x^2\right)-\left(6x^2-6x\right)-\left(x-1\right)\)

\(x^3\left(x-1\right)+6x^2\left(x-1\right)-6x\left(x-1\right)-\left(x-1\right)\)

\(\left(x-1\right)\left(x^3+6x^2-6x-1\right)\)

\(\left(x-1\right)\left\{\left(x^3-x^2\right)+\left(7x^2-7x\right)+\left(x-1\right)\right\}\)

\(\left(x-1\right)^2\left(x^2+7x+1\right)\)

\(\Delta=49-4=45\)

\(x1,2=\frac{-7+\sqrt{45}}{2},\frac{-7-\sqrt{45}}{2}\)

\(\left(x-1\right)^2\left(x-\frac{7+\sqrt{45}}{2}\right)\left(x-\frac{7-\sqrt{45}}{2}\right)\)

\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)

\(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)

\(=\left(6x+1-6x+1\right)^2\)

\(=4\)

\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

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

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

\(=-3x^3-3x\)

\(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=\left(3x^2+5x^2-8x^2\right)-\left(6x+5x\right)+24\)

\(=-11x+24\)