[-\(\frac{-2}{3}\)\(^{x^3}\)\(\left(2x-1\right)^m\)\(+\frac{2}{5}\)\(x^{m+3}\)] :\(\left(-\frac{2}{3}x^3\right)\)=0
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VM
14 tháng 2 2020
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
VM
7 tháng 10 2019
Bài 1:
b) \(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-\frac{1}{2}\\x=0+\frac{3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{4}\right\}.\)
c) \(\left(2x-5\right)^4=81\)
\(\Rightarrow2x-5=\pm3\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3+5=8\\2x=\left(-3\right)+5=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8:2\\x=2:2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{4;1\right\}.\)
d) \(3^{x+1}+3^{x+3}=810\)
\(\Rightarrow3^x.3^1+3^x.3^3=810\)
\(\Rightarrow3^x.\left(3^1+3^3\right)=810\)
\(\Rightarrow3^x.30=810\)
\(\Rightarrow3^x=810:30\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
6 tháng 7 2019
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
9 tháng 2 2020
Câu 1 :
- Gọi chiều dài miếng đất là x ( m, x > 6 )
=> Chiều rộng miếng đất là : x - 6 ( m )
=> Chu vi miếng đất đó là : \(2\left(x+x-6\right)\) ( m )
Theo đề bài chu vi mảnh đất đó là 60m nên ta có phương trình :
\(2\left(x+x-6\right)=60\)
=> \(2x-6=30\)
=> \(2x=24\)
=> \(x=12\) ( TM )
Mà diện tích mảnh đất là : \(x\left(x-6\right)\)
=> Smảnh đất = \(12\left(12-6\right)=12.6=72\left(m^2\right)\)
8 tháng 1 2020
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
26 tháng 3 2017
a) ta có :x2+2x+2=(x+1)2+1>0,với mọi x
x2+2x+3=(x+1)2+2>0,với mọi x
ĐKXĐ:x\(\in\)R.Đặt x2+2x+2=a (a>0),ta có:\(\dfrac{a-1}{a}+\dfrac{a}{a+1}=\dfrac{7}{6}\)
<=>\(\dfrac{6\left(a-1\right)\left(a+1\right)}{6a\left(a+1\right)}+\dfrac{6a^2}{6a\left(a+1\right)}=\dfrac{7a\left(a+1\right)}{6a\left(a+1\right)}\)
=>6(a2-1)+6a2=7a2+7a<=>6a2-6+6a2=7a2+7a<=>12a2-7a2-7a-6=0
<=>5a2-7a-6=0<=>(a-2)(5a+3)=0<=>a-2=0(vì a>0,nên 5a+3>0)
<=>a=2=>x2+2x+2=2<=>x(x+2)=0<=>\(|^{x=0}_{x+2=0< =>x=-2}\)
Vậy tặp nghiệm của PT là S\(=\left\{0;-2\right\}\)