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31 tháng 10 2021
\(2,\\ a,=2x^2+4x-3x-6-2x^2-4x-2=-3x-8\\ b,=\left[x-2+2\left(x+1\right)\right]^2=\left(x-2+2x+2\right)^2=9x^2\)
9P
1
9 tháng 1 2022
Bài 5:
Xét ΔBAC có
FG//AC
nên \(\dfrac{FG}{AC}=\dfrac{BG}{BC}=\dfrac{1}{2}\)
hay AC=16(m)
NP
2
28 tháng 1 2023
a: Khi x=3 thì \(A=\dfrac{3\cdot3}{3-2}=9\)
b: C=A+B
\(=\dfrac{3x}{x-2}-\dfrac{6}{x-2}-\dfrac{x^2+4x+4}{x^2-4}\)
\(=\dfrac{3x-6}{x-2}-\dfrac{x+2}{x-2}\)
\(=\dfrac{3x-6-x-2}{x-2}=\dfrac{2x-8}{x-2}\)
c: Để C nguyên thì 2x-4-4 chia hết cho x-2
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6\right\}\)
J
3
TT
28 tháng 6 2023
\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(=x-0,2-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(=\left(-0,2-2+2\right)+\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)\)
\(=-0,2\)
\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(=x^3-8y^3-x^3+8y^3-10\)
\(=-10\)
\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)
\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=13\)
28 tháng 6 2023
a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(A=x-\dfrac{1}{5}-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(A=\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)-\left(\dfrac{1}{5}+2-2\right)\)
\(A=-\dfrac{1}{5}\)
Vậy: ...
b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)
\(B=-10\)
Vậy: ...
c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)
\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)
\(=13\)
Vậy:...
6 tháng 2 2021
a) Ta có: \(\left(2x-3\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};-\dfrac{4}{3}\right\}\)
b) Ta có: \(x^3-3x^2+3x-1=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)^3-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2-2x+1-x-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-3x\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)
Vậy: S={0;1;3}
c) Ta có: \(x^2+x=2x+2\)
\(\Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy: S={-1;2}
d) Ta có: \(\left(x-1\right)^2=2\left(x^2-1\right)\)
\(\Leftrightarrow\left(x-1\right)^2-2\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1-2x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)Vậy: S={1;-3}
e) Ta có: \(2\left(x+2\right)^2-x^3-8=0\)
\(\Leftrightarrow2\left(x+2\right)^2-\left(x^3+8\right)=0\)
\(\Leftrightarrow2\left(x+2\right)\cdot\left(x+2\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x+4-x^2+2x-4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\cdot\left(-x^2+4x\right)=0\)
\(\Leftrightarrow-x\left(x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=4\end{matrix}\right.\)
Vậy: S={0;-2;4}
