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\(\left(x-2\right)^3+\left(5-2x\right)^3=0\)
\(\Leftrightarrow\left(x-2+5-2x\right)\left(\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(x^2-4x+4-\left(5x-4x^2-10+4x\right)+25-20x+4x^2\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(x^2-4x+4-5x+4x^2+10-4x+25-20x+4x^2\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(9x^2-33x+39\right)=0\)
Phân tích tiếp nhé
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
a)\(2x^3=x^2+2x-1\Leftrightarrow2x^3-x^2-2x+1=0\Leftrightarrow x^2\left(2x-1\right)-\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-1\right)=0\Leftrightarrow\left(2x-1\right)\left(x-1\right)\left(x+1\right)=0\)
<=> 2x-1=0 hoặc x-1=0 hoặc x+1=0 <=> x=1/2 hoặc x=1 hoặc x=-1
b)\(x^2-4+\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\5-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
\(giải:\)
\(1,\)\(\frac{x}{5}+\frac{2x+1}{3}=\frac{x-5}{15}\)
\(\Leftrightarrow\frac{x}{5}+\frac{2x+1}{3}-\frac{x-15}{15}=0\)
\(\Leftrightarrow\frac{3x}{15}+\frac{5\left(2x+1\right)}{15}-\frac{x-15}{15}=0\)
\(\Leftrightarrow\frac{3x+5\left(2x+1\right)-\left(x-15\right)}{15}=0\)
\(\Leftrightarrow\frac{3x+10x+5-x+15}{15}=0\)
\(\Leftrightarrow\frac{12x+20}{15}=0\)
\(\Rightarrow12x+20=0\)
\(\Leftrightarrow12x=-20\Leftrightarrow x=\frac{-5}{3}\)
vậy tập nghiệm của phương trình là \(s=\left[\frac{-5}{3}\right]\)
\(2,\)\(\left(x^3-64\right)+6x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-4^3\right)+6x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+16\right)+6x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+16+6x\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+10x+16\right)=0\)
\(mà\)\(x^2+10x+16>0\)
\(\Rightarrow x-4=0\Rightarrow x=4\)
vậy x=4 là nghiệm của phương trình
\(3,\)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{16}{x^2-4}\)
\(\Leftrightarrow\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{16}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{16}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\left(x+2\right)\left(x+2\right)-\left(x-2\right)\left(x-2\right)=16\)\
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2-16=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4-16=0\)
\(\Leftrightarrow8x-16=0\)
\(\Leftrightarrow8\left(x-2\right)=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
vậy x=2 là nghiệm của phương trình
a) Đặt: x = a- b; y = b - c ; z = c- a
Ta có: x + y + z = 0
=> \(A=x^3+y^3+z^3=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=3xyz\)
=> \(A=3xyz=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
b) Đặt: \(a=x^2-2x\)
Ta có: \(B=a\left(a-1\right)-6=a^2-a-6=\left(a+2\right)\left(a-3\right)=\left(x^2-2x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x^2-2x+2\right)\left(x+1\right)\left(x-3\right)\)
d) \(D=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+25x^2\)
Đặt: \(x^2-8=t\)
Ta có: \(D=4\left(t+2x\right)\left(t+7x\right)+25x^2\)
\(=4t^2+36xt+81x^2=\left(2t+9x\right)^2\)
\(=\left(2x^2+9x-16\right)^2\)
\(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
\(2x\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(2x-x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)\)
\(\left(x-2\right)^3+\left(5-2x\right)^3=0\)
\(\Leftrightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\)
\(\Leftrightarrow\left(3-x\right)\left(x^2-4x+4+2x^2-9x+10+4x^2-20x+25\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(7x^2-33x+39\right)=0\)
\(\Leftrightarrow3-x=0\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
Wish you study well !!
Bạn ơi, bạn có thể c/m cho mình \(7x^2-33x+39>0\) đc ko bạn?