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Khách
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JH
0
JH
1
18 tháng 8 2021
1: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
2: Ta có: \(\left(5x-4\right)^2-49x^2=0\)
\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(2x+4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3: Ta có: \(5x^3-20x=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Y
7 tháng 7 2023
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)
\(10,\left(x+3\right)^2-x^2=45\)
\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)
Vậy \(S=\left\{6\right\}\)
\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)
\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)
YN
2 tháng 10 2021
Bài 2: Tính giá trị của biểu thức sau:
\(16x^2-y^2=\left(4x+y\right)\left(4x-y\right)\)
Thay \(\hept{\begin{cases}x=87\\y=13\end{cases}}\)
\(\Rightarrow\left(4.87+13\right)\left(4.87-13\right)=361.335=120935\)
YN
2 tháng 10 2021
Bài 4: Tìm x
a) \(9x^2+x=0\)
\(\Rightarrow x\left(9x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\9x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{9}\end{cases}}\)
b) \(27x^3+x=0\)
\(\Rightarrow x\left(27x^2+1=0\right)\)
\(\Rightarrow\orbr{\begin{cases}x=0\\27x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\27x^2=\left(-1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{-1}{27}\end{cases}}\)
Ta có: \(\frac{-1}{27}\) loại vì \(x^2\ge0\forall x\)
Vậy \(x=0\)
T
1
6 tháng 11 2021
\(a,\Leftrightarrow\left(x-9\right)^2-2\left(x-9\right)+1=0\\ \Leftrightarrow\left(x-9-1\right)^2=0\Leftrightarrow x=10\\ b,Sửa:49x^2-14x\sqrt{5}+5=0\\ \Leftrightarrow\left(7x-\sqrt{5}\right)^2=0\Leftrightarrow x=\dfrac{\sqrt{5}}{7}\)
MR
1
KT
22 tháng 7 2018
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
11 tháng 8 2017
a. \(\left(3x-5\right)^2-\left(x+1\right)^2=0\Leftrightarrow\left(3x-5+x+1\right)\left(3x-5-x-1\right)=0\Leftrightarrow\left(4x-4\right)\left(2x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}4x-4=0\\2x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy ...
b. \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
c. \(4x^3-36x=0\Leftrightarrow4x\left(x^2-9\right)=0\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy ...
d. \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\Leftrightarrow\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-2x+3\right)=0\Leftrightarrow6\left(x-1\right)=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy ...
PN
10 tháng 5 2021
b, \(15\left(x+3\right)+20x\left(x+8\right)=15x+45+20x^2+160x\)
\(=20x^2+175x+45=...\)
c, \(6\left(x-9\right)-3x\left(y-x\right)=6x-54-3xy+3x^2\)
d, \(2xy+10x^2-x\) không phân tích được nữa nhé
e, \(4ab^2-28a+16b\)không phân tích được nữa nhé
g, \(a\left(a+b\right)=ab\left(a+b\right)< =>\left(a+b\right)\left(a-ab\right)=0< =>\left(a+b\right)a\left(1-b\right)=0\)
h, \(30a^2+6a-6=\left(\sqrt{30}a\right)^2+2.\sqrt{30}.\frac{3}{\sqrt{30}}+\frac{3}{10}-\frac{63}{10}\)
\(=\left(\sqrt{30}a+\frac{3}{\sqrt{30}}\right)^2-\sqrt{\frac{63}{10}}^2=\left(\sqrt{30}a+\frac{3}{\sqrt{30}}-\sqrt{\frac{63}{10}}\right)\left(\sqrt{30}a+\frac{3}{\sqrt{30}}+\sqrt{\frac{63}{10}}\right)\)
T
1
a, \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left(3x+3\right)^2=0\Leftrightarrow\left(4x-3x-3\right)\left(4x+2x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)
b, \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow x=-2;x=\frac{1}{3}\)
c, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)