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26 tháng 11 2017

giải theo tham số m

26 tháng 11 2017

Ê ! Điểm SP là điểm gì vậy ? 😄

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 \(x\in\left\{\frac{1}{4}\right\}\)

23 tháng 11 2017

\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)

\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)

\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)

\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)

\(=-\left(2x+\frac{7}{5}\right)^m\)

đến đây thì mình chịu

AH
Akai Haruma
Giáo viên
25 tháng 11 2020

Lời giải:

a) ĐKXĐ:

\(\left\{\begin{matrix} 2x+10\neq 0\\ x\neq 0\\ 2x(x+5)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq -5\\ x\neq 0\end{matrix}\right.\)

b)

\(B=\frac{x(x^2+2x)}{x(2x+10)}+\frac{(2x+10)(x-5)}{x(2x+10)}+\frac{50-5x}{x(2x+10)}\)

\(=\frac{x^3+2x^2+2(x^2-25)+50-5x}{x(2x+10)}=\frac{x^3+4x^2-5x}{2x(x+5)}=\frac{x^2+4x-5}{2(x+5)}=\frac{(x-1)(x+5)}{2(x+5)}=\frac{x-1}{2}\)

Để $B=0\Leftrightarrow \frac{x-1}{2}=0\Leftrightarrow x=1$ (thỏa mãn)

Để $B=\frac{1}{4}\Leftrightarrow \frac{x-1}{2}=\frac{1}{4}$

$\Leftrightarrow x-1=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}$ (thỏa mãn)

17 tháng 8 2020

Lời giải:

a) ĐKXĐ:

\(\left\{\begin{matrix} 2x+10\neq 0\\ x\neq 0\\ 2x(x+5)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq -5\\ x\neq 0\end{matrix}\right.\)

b)

\(B=\frac{x(x^2+2x)}{x(2x+10)}+\frac{(2x+10)(x-5)}{x(2x+10)}+\frac{50-5x}{x(2x+10)}\)

\(=\frac{x^3+2x^2+2(x^2-25)+50-5x}{x(2x+10)}=\frac{x^3+4x^2-5x}{2x(x+5)}=\frac{x^2+4x-5}{2(x+5)}=\frac{(x-1)(x+5)}{2(x+5)}=\frac{x-1}{2}\)

Để $B=0\Leftrightarrow \frac{x-1}{2}=0\Leftrightarrow x=1$ (thỏa mãn)

Để $B=\frac{1}{4}\Leftrightarrow \frac{x-1}{2}=\frac{1}{4}$

$\Leftrightarrow x-1=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}$ (thỏa mãn)

29 tháng 3 2020

Bài 2 :

a, Ta có : \(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

=> \(A=\frac{x-5}{\left(x+5\right)\left(x-5\right)}+\frac{2\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

=> \(A=\frac{x-5+2\left(x+5\right)-2x-10}{\left(x-5\right)\left(x+5\right)}\)

=> \(A=\frac{x-5}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x+5}\)

b, - Thay A = -3 ta được phương trình \(\frac{1}{x+5}=-3\)

=> \(-3\left(x+5\right)=1\)

=> \(-3x-15=1\)

=> \(-3x=16\)

=> \(x=-\frac{16}{3}\)

- Thay x = \(-\frac{16}{3}\)vào phương trình trên ta được :

\(9.\left(-\frac{16}{3}\right)^2-42.\left(-\frac{16}{3}\right)+49=529\)

24 tháng 1 2017

d)

\(x\ne a,x\ne b\)

đặt \(\frac{x-a}{x-b}=t\Leftrightarrow t+\frac{1}{t}=2\Leftrightarrow\frac{t^2-2t+1}{t}=0\Rightarrow t=1\)

\(\frac{x-a}{x-b}=1\Leftrightarrow\frac{\left(x-a\right)-\left(x-b\right)}{x-b}=\frac{b-a}{x-b}=0\)

Vậy: \(a\ne b\) Pt vô nghiệm

a=b phương trinhg nghiệm với mọi x khác a, b

25 tháng 1 2017

cảm ơn bạn nha

1 tháng 3 2020

1, \(=\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}+\frac{1}{13}\right)}{7\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{7}\)

2, a, \(\Leftrightarrow\left(3x-2\right)^{10}-\left(3x-2\right)^6=0\)

\(\Leftrightarrow\left(3x-2\right)^6\left[\left(3x-2\right)^4-1\right]=0\)

TH1: (3x-2)^6=0 <=> 3x-2=0 <=> x=2/3

TH2: (3x-2)^4-1=0 <=> (3x-2)^4=1

<=> 3x-2 = 1 hoặc 3x-2=-1

<=>x=1 hoặc x=-1/3

Vậy x=2/3 hoặc x=1 hoặc x=-1/3

b, \(\Leftrightarrow\orbr{\begin{cases}2x^2-13=-5\\2x^2-13=5\end{cases}\Leftrightarrow\orbr{\begin{cases}2x^2=8\\2x^2=18\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\pm2\\x=\pm3\end{cases}}}\)

1 tháng 3 2020

3,a, \(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)+21}{n-4}=3+\frac{21}{n-4}\)

Để \(A\in Z\Leftrightarrow\frac{21}{n-4}\in Z\Leftrightarrow21⋮n-4\Leftrightarrow n-4\inƯ\left(21\right)\)

Ta có bảng

n-41-13-37-721-21
n537111-325-17

Vậy..

b, tương tự a

19 tháng 3 2020

b)\(\frac{x-1}{x+5}=\frac{6}{7}\)

\(\Leftrightarrow7x-7=6x+30\)

\(\Leftrightarrow7x-6x=30+7\)

\(\Leftrightarrow x=37\)