Tìm x
a, x\(^2\)-x(x+2)-1=2
b, x\(^2\)=(2x-1)\(^2\)
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ZH
1
5 tháng 10 2021
\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)
ZH
2
9 tháng 11 2021
\(a,\Leftrightarrow x^3-4x^2+4x=0\\ \Leftrightarrow x\left(x^2-4x+4\right)=0\\ \Leftrightarrow x\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ b,\Leftrightarrow4\left(x-1\right)=3x+6\left(2x-3\right)\\ \Leftrightarrow4x-4=3x+12x-18\\ \Leftrightarrow11x=14\Leftrightarrow x=\dfrac{14}{11}\)
NI
9 tháng 11 2021
a/ \(x^3-4x^2=-4x\)
\(\Leftrightarrow x^3-4x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b/ \(\dfrac{x-1}{3}=\dfrac{x}{4}+\dfrac{2x-3}{2}\)
\(\Leftrightarrow8\left(x-1\right)=6x+12\left(2x-3\right)\)
\(\Leftrightarrow8x-8=6x+24x-36\)
\(\Leftrightarrow8x-8=30x-36\)
\(\Leftrightarrow8x-30x=8-36\)
\(\Leftrightarrow-22x=-28\)
\(\Leftrightarrow x=\dfrac{14}{11}\)
ZH
1
17 tháng 11 2021
\(a,\Leftrightarrow\left(x^2-8x+16\right)-10=0\\ \Leftrightarrow\left(x-4\right)^2-10=0\\ \Leftrightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\\ b,\Leftrightarrow10\left(2x-1\right)+6x=9x\\ \Leftrightarrow20x-10-3x=0\\ \Leftrightarrow17x=10\Leftrightarrow x=\dfrac{10}{17}\)
ZH
0
9 tháng 11 2021
Bài 1:
\(a,=6x^2+19x-7-6x^3-4x^2+7x=-6x^3+2x^2+26x-7\\ b,B=26\cdot\left(63^2+63\cdot37+37^2\right):26+63\cdot37\\ =63^2+63\cdot37+37^2+63\cdot37\\ =\left(63+37\right)^2=100^2=10000\)
Bài 2:
\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=\left(x-y\right)\left(x+2\right)\\ c,=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)
BP
2
7 tháng 10 2021
b: Ta có: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(\Leftrightarrow\left(3x+4-2x-7\right)\left(3x+4+2x+7\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{5}\end{matrix}\right.\)
7 tháng 10 2021
c: ta có: \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(3x-6\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-6-x-2\right)\left(3x-6+x+2\right)=0\)
\(\Leftrightarrow\left(2x-8\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
21 tháng 9 2021
\(a,\Leftrightarrow\left(x-5\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
MN
2
6 tháng 11 2021
\(a,=6x^2-4x-x^2-4x-4=5x^2-8x-4\\ b,=x^3+8-2\left(1-x^2\right)=x^3+8-2+2x^2=x^3+2x^2+6\\ c,=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\\ =\left(2x+1-2x+1\right)^2=4\)
6 tháng 11 2021
Có thể giúp mình thực hiện cách chi tiết ko ạ ? Gv dạy mik ko hiểu mấy
\(x^2-x^2-2x-1-2=0\)
\(-2x-3=0\Leftrightarrow x=\dfrac{-2}{3}\)
\(\left(x-2x+1\right)\left(x+2x-1\right)=0\)
\(\left[{}\begin{matrix}-x+1=0\\3x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
a)\(x^2-x\left(x+2\right)-1=2\\ \Rightarrow x^2-x^2-2x-1=2\\ \Rightarrow-2x=3\\ \Rightarrow x=-\dfrac{3}{2}\)
b) \(x^2=\left(2x-1\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=2x-1\\x=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)