nhân chéo là đc:

3(x+2)=-4(x-5)

3x+6=-4x+20

3x+4x=20-6

7x     =14

 x      =2

Vậy x=2

= 10/3.x+ 67/4=-53/4

   10/3.x=-53/4-67/4

   10/3.x=-30

   x=-30:10/3

   x=-9

\(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)

\(\frac{10}{3}x+\frac{67}{4}=\frac{-53}{4}\)

\(\frac{10}{3}x=-30\)

\(x=-9\)

\(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)

\(\frac{10}{3}x+\frac{67}{4}=\frac{-53}{4}\)

\(\frac{10}{3}x=-30\)

\(x=-9\)

\(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)

\(\frac{10}{3}x=-13,25-16\frac{3}{4}\)

\(\frac{10}{3}x=-30\)

\(x=-9\)

\(-52+\frac{2}{3}x=-46\)

\(\frac{2}{3}x=-46+52\)

\(\frac{2}{3}x=6\)

\(x=6:\frac{2}{3}\)

\(x=9\)

x=9 nha bn

đúng nha

Happy new year!!

\(-52+\frac{2}{3}x=-46\)

\(\Rightarrow\frac{2}{3}x=-46-\left(-52\right)\)

\(\Rightarrow\frac{2}{3}x=6\)

\(\Rightarrow x=9\)

b) \(\left(2,4.x-36\right)\div1\frac{5}{7}=-1\)

\(\left(2,4.x-36\right)=-1.\frac{12}{7}\)

\(2,4.x-36=-\frac{12}{7}\)

\(2,4.x=-\frac{12}{7}+36\)

\(2,4.x=\frac{240}{7}\)

\(x=\frac{240}{7}\div2,4\)

\(x=\frac{100}{7}\)

a) \(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Rightarrow3.\left(x+2\right)=-4.\left(x-5\right)\)

\(\Rightarrow3x+6=-4x+20\)

\(\Rightarrow7x=14\)

\(\Rightarrow x=2\)

Câu c tương tự nhé. 

\(ĐKXĐ:x\ne\pm3\)

\(pt\Leftrightarrow\frac{\left(x+3\right)^2-\left(x-3\right)^2}{x^2-9}=\frac{17}{x^2-9}\)

\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=17\)

Tự dừng bấm Gửi tl

\(\Leftrightarrow x^2+6x+9-x^2+6x-9=17\)

\(\Leftrightarrow12x=17\Leftrightarrow x=\frac{17}{12}\)

\(\frac{x+3}{x-3}-\frac{17}{x^2-9}=\frac{x-3}{x+3}\)

\(\frac{x+3}{x-3}-\frac{17}{x^2-3^2}=\frac{x-3}{x+3}\)

\(\frac{x+3}{x-3}-\frac{17}{\left(x+3\right)\left(x-3\right)}=\frac{x-3}{x+3}\)

\(\left(x+3\right)^2-17=\left(x-3\right)^2\)

\(x^2+6x-8=x^2-6x+9\)

\(6x-8=-6x+9\)

\(6x=-6x+9+8\)

\(6x+6x=17\)

\(12x=17\Leftrightarrow x=\frac{17}{12}\)

\(4\left(x+1\right)^2=\sqrt{2\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow16\left(x+1\right)^4=2\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\left(x^2+3x+1\right)\left(7x^2+11x+7\right)=0\)

\(\sqrt{\frac{x+56}{16}+\sqrt{x-8}}=\frac{x}{8}\)

\(\Leftrightarrow2\sqrt{x+56+16\sqrt{x-8}}=x\)

\(\Leftrightarrow2\sqrt{\left(\sqrt{x-8}+8\right)^2}=x\)

\(\Leftrightarrow2\sqrt{x-8}+16=x\)

\(\Leftrightarrow x=24\)

a) Với \(x\ge0\)và \(x\ne1\)ta có:

\(P=\frac{10\sqrt{x}}{x+3\sqrt{x}-4}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}\)

\(=\frac{10\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{10\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\frac{10\sqrt{x}-\left(2x-5\sqrt{x}+3\right)-\left(x+5\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\frac{10\sqrt{x}-2x+5\sqrt{x}-3-x-5\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{-\left(3x-10\sqrt{x}+7\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\frac{-\left(\sqrt{x}-1\right)\left(3\sqrt{x}-7\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{-3\sqrt{x}+7}{\sqrt{x}+4}\)

b) \(P=\frac{-3\sqrt{x}+7}{\sqrt{x}+4}=\frac{-3\sqrt{x}-12+19}{\sqrt{x}+4}=\frac{-3\left(\sqrt{x}+4\right)+19}{\sqrt{x}+4}=-3+\frac{19}{\sqrt{x}+4}\)

Vì \(x\ge0\); \(x\ne1\)\(\Rightarrow\sqrt{x}+4\ge4\)

\(\Rightarrow\frac{19}{\sqrt{x}+4}\le\frac{19}{4}\)\(\Rightarrow P\le-3+\frac{19}{4}=\frac{7}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow x=0\)( thỏa mãn )

Vậy \(maxP=\frac{7}{4}\)\(\Leftrightarrow x=0\)

\(\frac{1}{2}x-\frac{1}{4}=\frac{-1}{2}\)

\(\frac{1}{2}x=\frac{-1}{2}+\frac{1}{4}\)

\(\frac{1}{2}x=\frac{-2+1}{4}\)

\(\frac{1}{2}x=\frac{-1}{4}\)

\(x=\frac{-1}{4}:\frac{1}{2}\)

\(x=\frac{-1}{4}.2\)

\(x=\frac{-1}{2}\)

Vậy .....................

\(\frac{1}{2}x-\frac{1}{4}=-\frac{1}{2}\)

\(\frac{1}{2}x=-\frac{1}{2}+\frac{1}{4}\)

\(\frac{1}{2}x=-\frac{2}{4}+\frac{1}{4}\)

\(\frac{1}{2}x=-\frac{1}{4}\)

\(x=-\frac{1}{4}:\frac{1}{2}\)

\(x=-\frac{1}{4}\cdot2\)

\(x=-\frac{1}{2}\)

Vậy .....

\(\frac{1}{2}x-\frac{1}{4}=-\frac{1}{2}\)

\(\frac{1}{2}x=-\frac{1}{2}+\frac{1}{4}\)

\(\frac{1}{2}x=-\frac{1}{4}\)

\(x=-\frac{1}{4}:\frac{1}{2}\)

\(x=-\frac{1}{2}\)