\(9\text{|}x+3\text{|}-4\text{|}x-4\text{|}=18-x+5.\) ( quy đồng) mẫu chung là 36

phá dấu bừa nhé

TH1 :  \(9\left(-x-3\right)-4\left(-x+4\right)=18-x+5\)

          \(-9x-27+4x-16=18-x+5\)

              \(-4x=18+5+27+16=66\)

                  \(x=\frac{66}{-4}\)

TH2:  \(9\left(x+3\right)-4\left(x-4\right)=18-x+5\) ( quy đồng ) mẫu chung là 36

        \(9x+27-4x+16=18-x+5\)

      \(6x=18+5-27-16=-20.\)

  \(x=-\frac{20}{6}\)

p/s làm bừa nhé đừng chửi

phá dấu trị tuyệt đối ra có bị làm sao k ?

mk dg can gap

a,\(\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2=90\)\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+2.\frac{x}{x+1}.\frac{x}{x-1}+\left(\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)

\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)\(\Leftrightarrow\left(\frac{x^2-x+x^2+x}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}=90\)

\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-90=0\)\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}-10\right)\left(\frac{2x^2}{x^2-1}+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2x^2}{x^2-1}=10\\\frac{2x^2}{x^2-1}=-9\end{cases}\Leftrightarrow......}\)

b,Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\Rightarrow ab=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x^2-4}{x^2-1}\)

Từ đó ta có phương trình:\(20a^2-5b^2+48ab=0\Leftrightarrow20a^2-2ab-5b^2+50ab=0\)

\(\Leftrightarrow2a\left(10a-b\right)+5b\left(10a-b\right)=0\Leftrightarrow\left(2a+5b\right)\left(10a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2a=-5b\\10a=b\end{cases}}\)

TH1:\(2a=-5b\Leftrightarrow\frac{2\left(x-2\right)}{x+1}=\frac{-5\left(x+2\right)}{x-1}\)\(\Rightarrow2\left(x-2\right)\left(x-1\right)=-5\left(x+2\right)\left(x+1\right)\)\(\Leftrightarrow2x^2-6x+4=-5x^2-15x-10\)\(\Leftrightarrow7x^2+9x+14=0\)

\(\Leftrightarrow7\left(x^2+\frac{9}{7}x+2\right)=0\Leftrightarrow7\left(x^2+2.\frac{9}{14}+\frac{81}{196}\right)+\frac{311}{28}=0\)

\(\Leftrightarrow7\left(x+\frac{9}{14}\right)^2+\frac{311}{28}=0\),vô lí
TH2:Tự làm nhé ,tương tự

a/ \(\Leftrightarrow tanx.tan\frac{\pi}{9}-1=tan\frac{\pi}{90}\left(tanx+tan\frac{\pi}{9}\right)\)

\(\Leftrightarrow\frac{tanx+tan\frac{\pi}{9}}{1-tanx.tan\frac{\pi}{9}}=-\frac{1}{tan\frac{\pi}{90}}\)

\(\Leftrightarrow tan\left(x+\frac{\pi}{9}\right)=tan\left(\frac{23\pi}{45}\right)\)

\(\Rightarrow x+\frac{\pi}{9}=\frac{23\pi}{45}+k\pi\)

\(\Rightarrow x=\frac{2\pi}{5}+k\pi\)

Do \(-2\pi< x< 2\pi\Rightarrow-2\pi< \frac{2\pi}{5}+k\pi< 2\pi\)

\(\Rightarrow k=\left\{-2;-1;0;1;2\right\}\)

\(\Rightarrow x=\left\{-\frac{8\pi}{5};-\frac{3\pi}{5};\frac{2\pi}{5};\frac{7\pi}{5};\frac{12\pi}{5}\right\}\)

b/

ĐKXĐ: \(cos2x\ne0\)

\(\Leftrightarrow tan^22x+1+tan^22x=7\)

\(\Leftrightarrow tan^22x=3\)

\(\Rightarrow\left[{}\begin{matrix}tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tan2x=tan60^0\\tan2x=tan\left(-60^0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=60^0+k180^0\\2x=-60^0+k180^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-30^0+k180^0\end{matrix}\right.\)

Bạn tự tìm nghiệm thuộc khoảng đã cho nhé

c/ ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow tan^3x+1+tan^2x+4\sqrt{3}\left(1+tanx\right)=8+7tanx\)

\(\Leftrightarrow tan^2x\left(1+tanx\right)+\left(4\sqrt{3}-7\right)\left(1+tanx\right)=0\)

\(\Leftrightarrow\left(tan^2x-7+4\sqrt{3}\right)\left(1+tanx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tan^2x=7-4\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=2-\sqrt{3}\\tanx=-2+\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tanx=tan\left(-\frac{\pi}{4}\right)\\tanx=tan\left(\frac{\pi}{12}\right)\\tanx=tan\left(-\frac{\pi}{12}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{12}+k\pi\\x=-\frac{\pi}{12}+k\pi\end{matrix}\right.\)

Bạn tự tìm x thuộc khoảng đã cho

b) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)

=> \(\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)

=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

=> \(\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)

=> x - 100 = 0

=> x = 100

a) \(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}+2\frac{14}{90}\)

= \(\left(74\frac{19}{35}+15\frac{16}{35}\right).\frac{7}{90}+2\frac{14}{90}\)

=  90 . 7/90 + 194/90

= 630/90 + 194/90

= 824/90 = 412/45

b) (-2/5 + 3/7) - (4/9 + 12/20 - 13/35) + 7/35

= -2/5 + 3/7 - 4/9 -  3/5 + 13/35 + 7/35

= (-2/5 - 3/5) + 3/7 - 4/9 + (13/35 + 7/35)

= -1 + 3/7 - 4/9 + 4/7

= -1 + (3/7 + 4/7) - 4/9

= -1 + 1 - 4/9 = -4/9

\(\frac{90}{x}-\frac{36}{x-6}=2\)               MTC = x (x-6)         ĐK\(\hept{\begin{cases}x\ne0\\x\ne6\end{cases}}\)

\(\frac{90\left(x-6\right)}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}=\frac{2x\left(x-6\right)}{x\left(x-6\right)}\)

\(\frac{90x-540}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}-\frac{2x^2-12x}{x\left(x-6\right)}=0\)

\(90x-540-36x-2x^2+12x=0\)

\(-2x^2+66x-540=0\)

\(-2x^2+36x+30x-540=0\)

\(-2x\left(x-18\right)+30\left(x-18\right)=0\)

\(\left(x-18\right)\left(-2x+30\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-18=0\\-2x+30=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=18\\x=15\end{cases}}\)

vậy.....

ĐKXĐ:  \(x\ne0;\)  \(x\ne6\)

         \(\frac{90}{x}-\frac{36}{x-6}=2\)

\(\Leftrightarrow\)\(\frac{90\left(x-6\right)}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}=2\)

\(\Leftrightarrow\)\(\frac{90x-540-36x}{x\left(x-6\right)}=2\)

\(\Leftrightarrow\)\(\frac{54x-540}{x\left(x-6\right)}=2\)

\(\Leftrightarrow\)\(54x-540=2x\left(x-6\right)\)

\(\Leftrightarrow\)\(27x-270=x\left(x-6\right)\)

mk lm đc có vậy thôi.  tham khảo nha

cảm ơn câu trả lời của song ngư đúng r mk cx lm ra vậy 

Ta có hệ pt :

\(\left\{{}\begin{matrix}\frac{y}{3}=\frac{x}{4}\\\frac{z}{9}=\frac{x}{4}\\7x-3y+2z=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\7x-\frac{3.3x}{4}+\frac{2.9x}{4}=90\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\28x-9x+18x=360\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\37x=360\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{4}.\frac{360}{37}\\z=\frac{9}{4}.\frac{360}{37}\\x=\frac{360}{37}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{270}{37}\\z=\frac{810}{37}\\x=\frac{360}{37}\end{matrix}\right.\)

Vậy . . . . . . . . .