\(a\text{)}.\:\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\\ =x^4+4x^2+4-\left(x^2-4\right)\left(x^2+4\right)\\ =x^4+4x^2+4-x^4+16\\ =4x^2+20\)

\(b\text{)}.\:\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\\ =\left(x+1+x-1\right)\left(x+1-x+1\right)-3\left(x^2-1\right)\\ =4x-3x^2+3\)

ta có: \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Rightarrow\left[{}\begin{matrix}x+2=x+4\left(vô\:lí\right)\\\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(•5^x+5^{x+1}+5^{x-2}=151\\ 5^x\left(1+5+\dfrac{1}{25}\right)=151\\ 5^x=25\\ \Rightarrow x=2\)

\(•5^{x-1}+5^{x-2}+5^{x-3}=155\\ 5^x.\left(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}\right)=155\\ 5^x=625\\ \Rightarrow x=4\)

\(•5^{2+x}+5^{3+x}=750\\ 5^x\left(25+125\right)=750\\ 5^x=5\\ \Rightarrow x=1\)

sửa đề câu a \(x\left(y-z\right)+y\left(z-x\right)+z\left(x-y\right)\)\(a\text{)}\: x\left(y-z\right)+y\left(z-x\right)+z\left(x-y\right)\\ =xy-xz+yz-xy+zx-zy=0\)

câu b tương tự.

\(•\left(x^2-1\right)^2+1=x^2\\ \left(x^2-1\right)^2-x^2+1=0\\ x^4-2x^2+1-x^2+1=0\\ x^4-x^2-2x^2+2=0\\ \left(x^2-1\right)\left(x^2-2\right)=0\\ \left(x+1\right)\left(x-1\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\\x+\sqrt{2}=0\\x-\sqrt{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=-\sqrt{2}\\x=\sqrt{2}\end{matrix}\right.\)

\(\left(x+y+z\right)^3-x^3-y^3-z^3\\ =x^3+y^3+z^3-x^3-y^3-z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\\ =3\left(x+y\right)\left(y+z\right)\left(z+x\right)\:\left(đpcm\right)\)

\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\\ \dfrac{x+3}{2007}-\dfrac{x+3}{2008}-\dfrac{x+3}{2010}+\dfrac{x+3}{2009}=0\\ \left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x+3=0\Rightarrow x=-3\)

\(\left(\dfrac{3}{4}x-1\right)^3=\dfrac{1}{64}\\ \Rightarrow\dfrac{3}{4}x-1=\dfrac{1}{4}\\ \dfrac{3}{4}x=\dfrac{5}{4}\\ x=\dfrac{\dfrac{5}{4}}{\dfrac{3}{4}}=\dfrac{5}{3}\)

\(x^{15}=x\\ x^{15}-x=0\\ x\left(x^{14}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{14}=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\end{matrix}\right.\)

vậy ......

ta biết được tổng 3 góc trong 1 tam giác bằng 180 độ

gọi số đo của các góc A,B,C lần lượt là x,y,z

theo bài ra ta có:

x/2=y/5=z/9 và x+y+z=180

áp dụng ..... ta có:

\(\frac{x}{2}=\frac{y}{5}=\frac{z}{9}=\frac{x+y+z}{2+5+9}=\frac{180}{16}=\frac{45}{4}\)

từ x/2=45/4=>x=45/4.2=45/2=22,5

y/5=45/4=>y=45/4.5=225/4=56,25

z/9=45/4=>z/405/4=101,25

vậy ....