\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3

a, \(\dfrac{10-2x}{2}=\dfrac{25-5x}{5}\)

\(\Leftrightarrow\dfrac{2\left(5-x\right)}{2}=\dfrac{5\left(5-x\right)}{5}\)

\(\Leftrightarrow5-x=5-x\)

\(\Leftrightarrow0x=0\)

⇒ Có vô số giá trị của x thỏa mãn.

Vậy...

b, ĐKXĐ: \(x\ne\pm1\)

\(\dfrac{x-3}{x-1}-\dfrac{2x+1}{x+1}=\dfrac{x-x^2}{x^2-1}\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+1\right)-\left(2x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-x^2}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow x^2-2x-3-2x^2+x+1=x-x^2\)

\(\Leftrightarrow-2x=2\)

\(\Leftrightarrow x=-1\left(ktm\right)\)

Vậy...

a) Ta có: \(\dfrac{10-2x}{2}=\dfrac{25-5x}{5}\)

\(\Leftrightarrow5\left(10-2x\right)=2\left(25-5x\right)\)

\(\Leftrightarrow50-10x=50-10x\)

\(\Leftrightarrow0x=0\)(phương trình có vô số nghiệm)

Vậy: S={x|\(x\in R\)}

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)

a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

\(C=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)

a: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)

\(\Leftrightarrow20x^2-12x+15x-5< 20x^2+30x-100\)

=>3x-5<=30x-100

=>30x-100>3x-5

=>27x>95

hay x>95/27

b: \(\Leftrightarrow4\left(5x-2\right)-6\left(2x^2-x\right)< 4x\left(1-3x\right)-15x\)

\(\Leftrightarrow20x-8-12x^2+6x< 4x-12x^2-15x\)

=>26x-8<-11x

=>37x<8

hay x<8/37

a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10

=1/2x^5-3x^4-5x^3

b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x

=-15x^5+12x^4-9x^3+9x^2

c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)

d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)

Bạn đăng từng câu 1 nhé

a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b: \(=\dfrac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)

\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

1.

\(\Leftrightarrow2sinx.cosx+2cosx=0\)

\(\Leftrightarrow2cosx\left(sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-1\end{matrix}\right.\)

\(\Leftrightarrow cosx=0\) (do \(cosx=0\Leftrightarrow sinx=\pm1\) bao hàm luôn cả pt \(sinx=-1\))

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

2.

\(\Leftrightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+n360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+n180^0\end{matrix}\right.\)

Do \(-120^0< x< 90^0\Rightarrow\left\{{}\begin{matrix}-120^0< 35^0+k180^0< 90^0\\-120^0< 65^0+n180^0< 90^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\\n=\left\{-1;0\right\}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=35^0\\x=-115^0\\x=65^0\end{matrix}\right.\)

3. Làm tương tự câu 2

4.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(10x+\dfrac{4\pi}{5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{x}{2}-2\pi\right)\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}-2\pi\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)=-cos\left(\dfrac{x}{2}\right)=cos\left(\pi-\dfrac{x}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}10x+\dfrac{4\pi}{5}=\pi-\dfrac{x}{2}+k2\pi\\10x+\dfrac{4\pi}{5}=\dfrac{x}{2}-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)