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\(\left(x^2+2x\right)^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+4x^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\Leftrightarrow\left(x-1\right)\left(x+1\right)^2\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)
Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=1\end{matrix}\right.\)
e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
e)x3-4x+14x(x-2)=0
⇔ x(x2-4)+14x(x-2)=0
⇔ x(x-2)(x+2)+14x(x-2)=0
⇔ (x-2)(x2+2x+14x)=0
⇔ x(x-2)(x+16)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)
g)x2(x+1)-x(x+1)+x(x-1)=0
⇔ (x+1)(x2-x)+x(x-1)=0
⇔ x(x+1)(x-1)+x(x-1)=0
⇔ x(x-1)(x+2)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
\(M=\left(\dfrac{1}{3}t\right)^2-4\left(t-v\right)^2+2tv+9v^2\)
\(=\left(\dfrac{1}{3}\cdot6\right)^2-4\cdot\left(6+1\right)^2+2\cdot6\cdot\left(-1\right)+9\)
\(=4-28-12+9\)
=-27
\(N=8\left(x-3\right)\left(2x+3\right)+\left(2x-6\right)^2+4\left(2x+3\right)^2\)
\(=8\left(2x^2+3x-6x-9\right)+4x^2-24x+36+4\left(4x^2+12x+9\right)\)
\(=8\left(2x^2-3x-9\right)+4x^2-24x+36+16x^2+48x+36\)
\(=16x^2-24x-9+20x^2+24x+72\)
\(=36x^2\)
\(=36\cdot\dfrac{9}{4}=81\)
a. \(M=\left(\dfrac{t}{3}\right)^2+2tv+9v^2-4\left(t-v\right)^2\)
\(=\left(\dfrac{t}{3}+3v\right)^2-4\left(t-v\right)^2\)
\(=\left(\dfrac{t}{3}+3v-2t+2v\right)\left(\dfrac{t}{3}+3v+2t-2v\right)\)
\(=\left(\dfrac{t}{3}+5v-2t\right)\left(\dfrac{t}{3}+v+2t\right)\)
Thay \(t=6\) và \(v=-1\) vào \(M\), ta được
\(M=\left(2-5-12\right)\left(2-1+12\right)=-15.13=-195\)
a: \(M=m^2\left(m+n\right)-n^2m-n^3\)
\(=m^2\left(m+n\right)-n^2\left(m+n\right)\)
\(=\left(m+n\right)^2\left(m-n\right)\)
\(=\left(-2017+2017\right)^2\cdot\left(-2017-2017\right)\)
=0
b: \(N=n^3-3n^2-n\left(3-n\right)\)
\(=n^2\left(n-3\right)+n\left(n-3\right)\)
\(=n\left(n-3\right)\left(n+1\right)\)
\(=13\cdot10\cdot14=1820\)
a: Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left[2\left(x-2\right)^2+1\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
c: Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^3=0\)
\(\Leftrightarrow\left(x-1.5\right)^6-2\left(x-1.5\right)^3=0\)
\(\Leftrightarrow\left(x-1.5\right)^3\cdot\left[\left(x-1.5\right)^3-2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1.5\\x=\sqrt[3]{2}+1.5\end{matrix}\right.\)
bài a) bn trên đã dẫn link cho bn r
bài b)
Đặt x-y=a;y-z=b;z-x=c
\(=>a+b+c=x-y+y-z+z-x=0\)
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)
Theo câu a)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) (do a+b+c=0)
\(=>a^3+b^3+c^3=3abc=>\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
a) Ta có :
\(a^3+b^3+c^3-3abc\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b^2\right)-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
P/s tham khảo nha
hok tốt
Ta có: \(\left(x-2\right)^3+\left(5-2x\right)^3=0\)
\(\Leftrightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\)
\(\Leftrightarrow3-x=0\)
hay x=3