\(a+b=5\Leftrightarrow a^2+2ab+b^2=25\)

\(\Leftrightarrow a^2+2.4+b^2=25\Leftrightarrow a^2+b^2=17\)

\(\Leftrightarrow\left(a^2+b^2\right)^2=289\Leftrightarrow a^4+2\left(ab\right)^2+b^4=289\)

\(\Leftrightarrow a^4+b^4+2.4^2=289\Leftrightarrow a^4+b^4=257\)

tích đi

=31

có thằng Phạm Huy đấy , bảo nó ở bên 

#Girl2k6#

    \(\frac{2x+1}{3x+1}=-1\)

\(\Rightarrow2x+1=-3x-1\)

\(\Rightarrow2x+3x=-1-1\)

\(\Rightarrow5x=-2\Rightarrow x=-\frac{2}{5}\)

\(\frac{2x+1}{3x+1}=-1\)

\(\Leftrightarrow2x+1=-3x-1\)

\(\Leftrightarrow5x=-2\)

\(\Leftrightarrow x=-\frac{2}{5}\)

\(\frac{2xy}{x^2-y^2}+\frac{x-y}{2x+2y}\)

\(=\frac{2xy}{\left(x-y\right)\left(x+y\right)}+\frac{x-y}{2\left(x+y\right)}\)

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

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

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

\(=\frac{\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}\)

\(=\frac{x-y}{2\left(x+y\right)}=\frac{x-y}{2x+2y}\)

x=0 hoặc x=1

k mk nhé

Trả lời 

đẻ x²=x³

=> x = 0 hoặc x = 1

hc tốt

Đặt biểu thức trên là A
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\ne0\)

\(\Rightarrow x=ak,y=bk,z=ck\)

Nên \(A=\frac{\text{[}\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2\text{]}.\left(a^2+b^2+c^2\right)}{\left(a.ak+b.bk+c.bk\right)^2}\)

\(=\frac{\left(a^2k^2+b^2k^2+c^2k^2\right).\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}\)
\(=\frac{k^2\left(a^2+b^2+c^2\right).\left(a^2+b^2+c^2\right)}{\text{[}k\left(a^2+b^2+c^2\right)\text{]}^2}\)

\(=\frac{k^2.\left(a^2+b^2+c^2\right)^2}{k^2.\left(a^2+b^2+c^2\right)}\)

\(=1\)

Vậy A=1

à quên sửa dòng trên chỗ A=1 cái chỗ mẫu là \(k^2.\left(a^2+b^2+c^2\right)^2\)nhen :v

\(x^4+x^3+2x^2+x+1\)

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

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

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

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

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