Giải phương trình sau:
c) x^2 -x-20=0
d) 2x^3+x-2x^2=1
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5 tháng 2 2022
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
28 tháng 8 2021
a: Ta có: \(6-4x=5(x+3)+3\)
\(\Leftrightarrow6-4x-5x-12-3=0\)
\(\Leftrightarrow-9x=9\)
hay x=-1
b: Ta có: \(\dfrac{x+3}{2}-1=\dfrac{x-1}{3}+\dfrac{x+5}{6}\)
\(\Leftrightarrow15x+45-30=10x-30+5x+25\)
\(\Leftrightarrow15=-5\left(loại\right)\)
c: Ta có: \(\left(x-2\right)\left(2x+1\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
29 tháng 8 2021
d: Ta có: \(\dfrac{2}{x^2-2x}+\dfrac{1}{x}=\dfrac{x+2}{x-2}\)
\(\Leftrightarrow2+x-2=x^2+2x\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
14 tháng 2 2022
a: =>x-3=2 hoặc x-3=-2
=>x=5 hoặc x=1
b: =>x2=0
hay x=0
c: =>(3x-5-x+1)(3x-5+x-1)=0
=>(2x-4)(4x-6)=0
=>x=2 hoặc x=3/2
d: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x-1-x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-4\right)=0\)
hay \(x\in\left\{1;-1;4\right\}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
14 tháng 2 2022
\(a,\left(x-3\right)^2=4\\\Leftrightarrow\left(x-3\right)^2-2^2=0\\ \Leftrightarrow \left(x-3-2\right).\left(x-3+2\right)=0\\ \Leftrightarrow\left(x-5\right).\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\\\Rightarrow S=\left\{1;5\right\}\\ b,x^2.\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\\ \Rightarrow S=\left\{0\right\}\\ c,\left(3x-5\right)^2-\left(x-1\right)^2=0\\ \Leftrightarrow\left(3x-5-x+1\right).\left(3x-5+x-1\right)=0\\ \Leftrightarrow\left(2x-4\right).\left(4x-6\right)=0\\ \Leftrightarrow2.\left(x-2\right).2.\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow S=\left\{\dfrac{3}{2};2\right\}\)
\(d,\left(x^2-1\right).\left(2x-1\right)=\left(x^2-1\right).\left(x+3\right)\\ \Leftrightarrow\left(x^2-1\right).\left(2x-1-x-3\right)=0\\ \Leftrightarrow\left(x^2-1\right).\left(x-4\right)=0\\ \Leftrightarrow\left(x-1\right).\left(x+1\right).\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=4\end{matrix}\right.\\ \Rightarrow S=\left\{-1;1;4\right\}\)
6 tháng 1 2023
b)x+3=4:2
=> x=-1
d)5x-15=3x-5
<=> 5x-3x=15-5
<=> 2x=10
<=> x=5
f) 35-7x=11-5x
<=> 35-11=-5x+7x
<=> 24=2x
<=> x=12
6 tháng 1 2023
h) 6x-2-3x=10
<=> 3x=10+2
<=> x=4
j)3-2x=3x+3-x-2
<=> 3-2x=2x+1
<=>-4x=-2
<=> x=1/2
23 tháng 2 2021
tham khảo
https://hoidapvietjack.com/q/57243/giai-cac-phuong-trinh-sau-a-2x12-2x-12-b-x2-3x-2-5x2-3x60
TM
23 tháng 2 2021
b) (2x+1)2-2x-1=2
\(< =>4x^2+4x+1-2x-1=2\)
\(< =>4x^2+2x-2=0\)
\(< =>4x^2+4x-2x-2=0\)
\(< =>\left(4x^2+4x\right)-\left(2x+2\right)=0\)
\(< =>4x\left(x+1\right)-2\left(x+1\right)=0\)
\(< =>\left(x+1\right)\left(4x-2\right)=0\)
\(=>\left\{{}\begin{matrix}x+1=0=>x=-1\\4x-2=0=>x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy....
4 tháng 1 2022
\(a,\Leftrightarrow\left(4-5x\right)\left(4+5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(3x+1-2x\right)\left(3x+1+2x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{5}\end{matrix}\right.\\ d,Sửa:\left(4x+1\right)^2-\left(x-2\right)^2=0\\ \Leftrightarrow\left(4x+1-x+2\right)\left(4x+1+x-2\right)=0\\ \Leftrightarrow\left(3x+3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{5}\end{matrix}\right.\\ e,\Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
HP
11 tháng 1 2022
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
11 tháng 1 2022
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
11 tháng 1 2022
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)
11 tháng 1 2022
c: =>(x-3)(x2+3x+5)=0
=>x-3=0
hay x=3
d: =>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x-3)(x-4)=0
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
Giải phương trình :
c) x^2 - x - 20 = 0
=> x\(^2\)- 5x + 4x - 20 = 0
=> ( x\(^2\)+ 4x ) - ( 5x + 20 ) = 0
=> x ( x + 4 ) - 5 ( x + 4 ) =0
=> ( x + 4 ) ( x - 5 ) = 0
<=> \(\orbr{\begin{cases}x+4=0\\x-5=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-4\\x=5\end{cases}}\)
Vậy tập no của phương trình là S = -4 ; 5
\(x^2-x-20=0\)
\(\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)-\frac{81}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{81}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{81}{4}\)
\(\orbr{\begin{cases}x-\frac{1}{2}=\frac{-9}{2}\\x-\frac{1}{2}=\frac{9}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=5\end{cases}}\)
Vậy phương trình có tập nghiệm \(S=\left\{-4;5\right\}\)