Bài 1: tim x
\(\left(x-1\right)^{40}=\left(x-1\right)^{42}\)
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TD
7 tháng 7 2017
a ) \(3x\left(x-1\right)-x\left(3x-2\right)=5\)
\(\Leftrightarrow3x^2-3x-3x^2+2x=5\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy phương trình có nghiệm x = - 5 .
1 tháng 6 2022
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
22 tháng 3 2021
\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)
\(\Rightarrow32x+32=31x+62\)
\(\Rightarrow x=30\)
Vậy x=30
Chúc bn học tốt
NQ
0
2
11 tháng 1 2023
Bài `1:`
`h)(3/4x-1)(5/3x+2)=0`
`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`
______________
Bài `2:`
`b)3x-15=2x(x-5)`
`<=>3(x-5)-2x(x-5)=0`
`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`
`d)x(x+6)-7x-42=0`
`<=>x(x+6)-7(x+6)=0`
`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`
`f)x^3-2x^2-(x-2)=0`
`<=>x^2(x-2)-(x-2)=0`
`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`
`h)(3x-1)(6x+1)=(x+7)(3x-1)`
`<=>18x^2+3x-6x-1=3x^2-x+21x-7`
`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`
`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`
`j)(2x-5)^2-(x+2)^2=0`
`<=>(2x-5-x-2)(2x-5+x+2)=0`
`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`
`w)x^2-x-12=0`
`<=>x^2-4x+3x-12=0`
`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`
2
11 tháng 1 2023
`m)(1-x)(5x+3)=(3x-7)(x-1)`
`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`
`<=>(1-x)(5x+3+3x-7)=0`
`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`
`p)(2x-1)^2-4=0`
`<=>(2x-1-2)(2x-1+2)=0`
`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`
`r)(2x-1)^2=49`
`<=>(2x-1-7)(2x-1+7)=0`
`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`
`t)(5x-3)^2-(4x-7)^2=0`
`<=>(5x-3-4x+7)(5x-3+4x-7)=0`
`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`
`u)x^2-10x+16=0`
`<=>x^2-8x-2x+16=0`
`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`
BC
1
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 1 2021
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)
Đặt \(x^2+x=t\)
\(\Rightarrow t\left(t+1\right)=42\)
\(\Leftrightarrow t^2+t-42=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x=6\\x^2+x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+7=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
22 tháng 10 2016
a) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2x+3\\x+\frac{1}{2}=-\left(2x+3\right)\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x-x=\frac{1}{2}-3\\x+\frac{1}{2}=-2x-3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x+2x=-3-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\3x=\frac{-7}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x=\frac{-7}{6}\end{array}\right.\)
Vậy \(x\in\left\{\frac{-5}{2};\frac{-7}{6}\right\}\)
LY
22 tháng 10 2016
\(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(Ta\) \(có\): \(x+\frac{1}{2}=2x+3\)
\(x+\frac{1}{2}=x+x+3\\\)
\(x+\frac{1}{2}=x+\left(x+3\right)\)
\(\Rightarrow\frac{1}{2}=x+3\)
\(\Rightarrow x=\frac{1}{2}-3\)
\(\Rightarrow x=-\frac{5}{2}\)
Vậy \(x=-\frac{5}{2}\)
b, \(\left|x+\frac{1}{5}\right|+\left|x+\frac{2}{5}\right|+\left|x+1\frac{2}{5}\right|=4x\)
\(Ta\) \(có\)
\(x+\frac{1}{5}+x+\frac{2}{5}+x+1\frac{2}{5}\)\(=4x\)
\(3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)
\(3x+2=4x\)
\(3x+2=3x+x\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
hình như bài này gần gióng vs bài lớp 6 (ý kiến riêng ) đừng ném dá
Ta có: \(\left(x-1\right)^{40}=\left(x-1\right)^{42}\)
\(\Leftrightarrow\left(x-1\right)^{42}-\left(x-1\right)^{40}=0\)
\(\Leftrightarrow\left(x-1\right)^{40}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^{40}=0\\\left(x-1\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2;x=0\end{cases}}\)
Vậy \(x\in\left\{0;1;2\right\}\)