giai pt:
3x-1. 22x-2 =129-x
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
V
0
26 tháng 3 2017
\(x^2-22.x-110=0\)
<=>\(x^2-22.x=110\)
<=> \(x^2-22.x+11^2=110+11^2\)( cộng cả hai vế với \(11^2\)để được hằng đẳng thức)
<=>\(\left(x-11\right)^2=231\)
<=>\(\hept{\begin{cases}x-11=\sqrt{231}\\x-11=-\sqrt{231}\end{cases}}\)
<=>\(\hept{\begin{cases}x=11+\sqrt{231}\\x=11-\sqrt{231}\end{cases}}\)
vậy phương trình có hai nghiệm \(x_1=11+\sqrt{231};x_2=11-\sqrt{231}\)
6 tháng 3 2019
(3x-1) (x2 +2) = (3x-1)(7x-10)
=> (3x-1) (x2+2)-(3x-1)(7x-10)=0
=>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x2-7x+12)=0
=>(3x-1)(x-3)(x-4)=0
=>3x-1=0 => x= 1/3
x-3=0 => x=3
x-4=0 => x=4
vậy pt có tập nghiệm S={ 1/3; 3; 4}
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 6 2019
Câu 1: ĐKXĐ: ...
\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)
\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)
\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)
\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)
\(\Leftrightarrow...\)
Câu 2:
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)
\(\Leftrightarrow x\left(-3y^2\right)=y^3+2y\)
\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)
\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)
\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)
\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)
\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)
28 tháng 6 2019
\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{4x\left(3x-1\right)+x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)
\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)
\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)
\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)
\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)
Tìm được mỗi nghiệm thôi à :v
22 tháng 3 2018
Giải pt :
\(\left(x+2\right)\left(3x+1\right)+x^2=4\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+x^2-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[\left(3x+1\right)+\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{4}\end{matrix}\right.\)
Tập nghiệm của pt là : \(S=\left\{-2;\dfrac{1}{4}\right\}\)
12 tháng 5 2020
Bài 1:
a) Ta có: 22x-13=x-6
\(\Leftrightarrow22x-13-x+6=0\)
\(\Leftrightarrow21x-7=0\)
\(\Leftrightarrow21x=7\)
hay \(x=\frac{1}{3}\)
Vậy: \(x=\frac{1}{3}\)
b) Ta có: (x-7)(2x+10)=0
\(\Leftrightarrow\left(x-7\right)\cdot2\cdot\left(x+5\right)=0\)
mà \(2\ne0\)
nên \(\left[{}\begin{matrix}x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;7\right\}\)
c) ĐKXĐ: \(x\ne14\)
Ta có: \(\frac{12x+9}{x-14}=7\)
\(\Leftrightarrow12x+9=7\left(x-14\right)\)
\(\Leftrightarrow12x+9=7x-98\)
\(\Leftrightarrow12x+9-7x+98=0\)
\(\Leftrightarrow5x+107=0\)
\(\Leftrightarrow5x=-107\)
hay \(x=\frac{-107}{5}\)(tm)
Vậy: \(x=\frac{-107}{5}\)
d) Ta có: \(\frac{x+2}{4}+\frac{3x-4}{6}=\frac{x-14}{24}\)
\(\Leftrightarrow\frac{6\left(x+2\right)}{24}+\frac{4\left(3x-4\right)}{24}=\frac{x-14}{24}\)
Suy ra: \(6\left(x+2\right)+4\left(3x-4\right)=x-14\)
\(\Leftrightarrow6x+12+12x-16-x+14=0\)
\(\Leftrightarrow17x+10=0\)
\(\Leftrightarrow17x=-10\)
hay \(x=\frac{-10}{17}\)
Vậy: \(x=\frac{-10}{17}\)
LT
1
10 tháng 7 2023
=>\(\dfrac{-1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}=2\)
=>\(\dfrac{1}{x-4}-\dfrac{1}{x-1}=2\)
=>\(\dfrac{x-1-x+4}{x^2-5x+4}=2\)
=>2x^2-10x+8=3
=>2x^2-10x+5=0
=>\(x=\dfrac{5\pm\sqrt{15}}{2}\)
\(pt\Leftrightarrow \dfrac{3^x}{3}.\dfrac{4^x}{4}=12^{9-x}\Leftrightarrow 12^{x-1}=12^{9-x}\)
Suy ra x-1=9-x nên x=5