Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b,\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)ĐKXĐ : \(x\ne2;\ne-3\)
\(\Leftrightarrow\frac{x^2-9}{\left(x-2\right)\left(x+3\right)}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow x^2-9=5\)
\(\Leftrightarrow x^2=14\)
\(x=\sqrt{14}\)
.....
a) \(\left(x+3\right)^2-\left(x-3\right)^2=6x\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=6x\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=6x\Leftrightarrow12x=6x\)\(\Leftrightarrow12x-6x=0\Leftrightarrow6x=0\Leftrightarrow x=0\)
Vậy phương trình có tập nghiệm S = { 0 }
b)\(-ĐKXĐ:\hept{\begin{cases}x-2\ne0\\\left(x-2\right)\left(x+3\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-3\end{cases}}\)
- Ta có : \(\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\Leftrightarrow\frac{x-3}{x-2}-\frac{5}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-5}{\left(x-2\right)\left(x+3\right)}=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\left(thoaman\right)\\x=-3\left(kothoaman\right)\end{cases}}\)
Vậy phương trình có tập nghiệm S = { 3 }
\(\frac{12x^2+30x-21}{16x^2-9}-\frac{3x-7}{3-4x}=\frac{6x+5}{4x+3}\)
ĐKXĐ: \(x\ne\pm\frac{3}{4}\)
\(< =>\frac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}+\frac{3x-7}{4x-3}=\frac{6x+5}{4x+3}\)
\(=>12x^2+30x-21+\left(3x-7\right)\left(4x+3\right)=\left(6x+5\right)\left(4x-3\right)\)
\(< =>12x^2+30x-21+12x^2-19x-21=24x^2+2x-15\)
\(< =>24x^2+11x-42=24x^2+2x-15\)
\(< =>24x^2+11x-42-24x^2-2x+15=0\)
\(< =>9x-27=0\)
\(< =>x=3\left(TM\right)\)
Tập nghiệm phương trình \(S=\left\{3\right\}\)
\(\frac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}\)-\(\frac{3x-7}{3-4x}\)=\(\frac{6x+5}{4x+3}\)
\(\frac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}\)+\(\frac{\left(3x-7\right)\left(4x+3\right)}{\left(4x-3\right)\left(4x+3\right)}\)=\(\frac{\left(6x+5\right)\left(4x-3\right)}{\left(4x-3\right)\left(4x+3\right)}\)
12x2+30x-21+12x2-28x+9x-21=24x2+20x-18x-15
12x2+12x2-24x2+30x-28x+9x-20x+18x=21+21-15
-9x =27
x =\(\frac{27}{-9}\)
x =-3
\(\dfrac{6x+5}{12x+9}+\dfrac{3x-7}{9-12x}=\dfrac{4x^2+10x-7}{16x^2-9}\)
\(\Leftrightarrow\dfrac{6x+5}{3\left(4x+3\right)}-\dfrac{3x-7}{3\left(4x-3\right)}=\dfrac{12x^2+30x-21}{3\left(4x-3\right)\left(4x+3\right)}\)
\(\Leftrightarrow\left(6x+5\right)\left(4x-3\right)-\left(3x-7\right)\left(4x+3\right)=12x^2+30x-21\)
\(\Leftrightarrow24x^2-18x+20x-15-\left(12x^2+9x-28x-21\right)=12x^2+30x-21\)
\(\Leftrightarrow24x^2+2x-15-12x^2+19x+21=12x^2+30x-21\)
=>31x+6=30x-21
=>x=-27
a)\(-ĐKXĐ:\hept{\begin{cases}x-14\ne0;x-13\ne0\\x-9\ne0\\x-11\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne14;x\ne13\\x\ne9\\x\ne11\end{cases}}\)
- Ta có : \(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)
\(\Leftrightarrow\frac{2}{x-14}-\frac{5}{x-13}-\frac{2}{x-9}+\frac{5}{x-11}=0\)
\(\Leftrightarrow\left(\frac{2}{x-14}-\frac{2}{x-9}\right)-\left(\frac{5}{x-13}-\frac{5}{x-11}\right)=0\)
\(\Leftrightarrow2\left(\frac{1}{x-14}-\frac{1}{x-9}\right)-5\left(\frac{1}{x-13}-\frac{1}{x-11}\right)=0\)\(\Leftrightarrow2.\frac{\left(x-9\right)-\left(x-14\right)}{\left(x-9\right)\left(x-14\right)}-5.\frac{\left(x-11\right)-\left(x-13\right)}{\left(x-11\right)\left(x-13\right)}=0\)
\(\Leftrightarrow2.\frac{5}{\left(x-9\right)\left(x-14\right)}-5.\frac{2}{\left(x-11\right)\left(x-13\right)}=0\)
\(\Leftrightarrow\frac{10}{\left(x-9\right)\left(x-14\right)}-\frac{10}{\left(x-11\right)\left(x-13\right)}=0\)
\(\Leftrightarrow10\left[\frac{1}{\left(x-9\right)\left(x-14\right)}-\frac{1}{\left(x-11\right)\left(x-13\right)}\right]=0\)
\(\Leftrightarrow\frac{\left(x-11\right)\left(x-13\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}-\frac{\left(x-9\right)\left(x-14\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}=\) \(0\)
\(\Leftrightarrow\left(x-11\right)\left(x-13\right)-\left(x-9\right)\left(x-14\right)=0\)
\(\Leftrightarrow x^2-24x+143-x^2+23x-126=0\)
\(\Leftrightarrow-x+17=0\Leftrightarrow-x=-17\Leftrightarrow x=17\)
Vậy pt có tập nghiệm S = { 17 }
P/s: Mk làm hơi lòng vòng, bn thông cảm nhé !
Đkxđ: ...
Đặt x2+ 15=a (a>0)
Pt ban đầu trở thành:
(a-10x)/(a-6x)=4x/(a-12x)
<=>a2-26ax+144x2=0
<=>(a-12x)(a-10x)=0
Xét th:a=10x
pt có nghiệm \(X=5\pm\sqrt{10}\)
Xét th:a=12x
Pt có nghiệm \(X=6\pm\sqrt{21}\)
\(\Leftrightarrow\frac{6x+5}{12x+9}-\frac{3x-7}{12x-9}=\frac{4x^2+10x-7}{16x^2-9}.\)
\(\Leftrightarrow\frac{\left(6x+5\right)\left(12x-9\right)-\left(3x-7\right)\left(12x+9\right)}{\left(3.4.x\right)^2-\left(3.3\right)^2}=\frac{4x^2+10x-7}{16x^2-9}\)
\(\Leftrightarrow\frac{72x^2+6x-45-\left(36x^2-57x-63\right)}{3^2\left(16x^2-9\right)}=\frac{4x^2+10x-7}{16x^2-9}\)
ĐK: \(16x^2-9\ne0\Leftrightarrow x^2\ne\left(\frac{3}{4}\right)^2\Rightarrow x\ne\pm\frac{3}{4}\)
\(\Leftrightarrow72x^2+6x-45-36x^2+57x+63=36x^2+90x-63\)
\(\Leftrightarrow27x=81\Leftrightarrow x=3\)