1:

c: =>1/3x+2/3-x+1>x+3

=>-2/3x+5/3-x-3>0

=>-5/3x-4/3>0

=>-5x-4>0

=>x<-4/5

d: =>3/2x+5/2-1<=1/3x+2/3+x

=>3/2x+3/2<=4/3x+2/3

=>1/6x<=2/3-3/2=-5/6

=>x<=-5

2:

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

\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{120}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)

\(\Leftrightarrow-24x+144=-5x+30\)

\(\Leftrightarrow-24x+5x=30-144\)

\(\Leftrightarrow-19x=-114\)

hay x=6

Vậy: S={6}

b) Ta có: \(\dfrac{4-5x}{6}=\dfrac{2\left(-x+1\right)}{2}\)

\(\Leftrightarrow2\cdot\left(4-5x\right)=12\left(-x+1\right)\)

\(\Leftrightarrow2-10x=-12x+12\)

\(\Leftrightarrow2-10x+12x-12=0\)

\(\Leftrightarrow2x-10=0\)

\(\Leftrightarrow2x=10\)

hay x=5

Vậy: S={5}

c) Ta có: \(\dfrac{-\left(x-3\right)}{2}-2=\dfrac{5\left(x+2\right)}{4}\)

\(\Leftrightarrow\dfrac{2\left(3-x\right)}{4}-\dfrac{8}{4}=\dfrac{5\left(x+2\right)}{4}\)

\(\Leftrightarrow6-2x-8=5x+10\)

\(\Leftrightarrow-2x+2-5x-10=0\)

\(\Leftrightarrow-7x-8=0\)

\(\Leftrightarrow-7x=8\)

hay \(x=-\dfrac{8}{7}\)

Vậy: \(S=\left\{-\dfrac{8}{7}\right\}\)

d) Ta có: \(\dfrac{7-3x}{2}-\dfrac{5+x}{5}=1\)

\(\Leftrightarrow\dfrac{5\left(7-3x\right)}{10}-\dfrac{2\left(x+5\right)}{10}=\dfrac{10}{10}\)

\(\Leftrightarrow35-15x-2x-10-10=0\)

\(\Leftrightarrow-17x+15=0\)

\(\Leftrightarrow-17x=-15\)

hay \(x=\dfrac{15}{17}\)

Vậy: \(S=\left\{\dfrac{15}{17}\right\}\)

a) Ta có: ⇔6(x+4)30−30x30+12030=10x30−15(x−2)30⇔6(x+4)30−30x30+12030=10x30−15(x−2)30

hay x=6

Vậy: S={6}

b) Ta có: 

Vậy: 7−3x2−5+x5=17−3x2−5+x5=1

Vậy: x+45−x+4=x3−x−22x+45−x+4=x3−x−22

hay x=5

Vậy: S={5}

c) Ta có: ⇔2(3−x)4−84=5(x+2)4⇔2(3−x)4−84=5(x+2)4

hay S={−87}S={−87}

d) Ta có: ⇔5(7−3x)10−2(x+5)10=1010⇔5(7−3x)10−2(x+5)10=1010

hay S={1517}

Sửa lại câu c) đặt \(\sqrt{x}+1=\)t \(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)

a) \(\left(\sqrt{4-3x}\right)^2=8^2\)\(\Leftrightarrow4-3x=64\Rightarrow x=-20\)

b) \(\sqrt{4x-8}+1=12\sqrt{\dfrac{x-2}{9}}\Leftrightarrow2\sqrt{x-2}+1\)\(=\left(12\sqrt{\left(x-2\right).\dfrac{1}{9}}\right)\)

\(\Leftrightarrow2t+1=12.\dfrac{1}{3}t\) (Đặt t = \(\sqrt{x-2}\))

\(\Rightarrow t=\dfrac{1}{2}\) \(\Rightarrow\sqrt{x-2}=\dfrac{1}{2}\)\(\Rightarrow x=\dfrac{9}{4}\)

c) pt\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)

a) Ta có: \(\sqrt{4-3x}=8\)

\(\Leftrightarrow4-3x=64\)

\(\Leftrightarrow3x=4-64=-60\)

hay x=-20

b) Ta có: \(\sqrt{4x-8}-12\cdot\sqrt{\dfrac{x-2}{9}}=-1\)

\(\Leftrightarrow2\cdot\sqrt{x-2}-12\cdot\dfrac{\sqrt{x-2}}{3}=-1\)

\(\Leftrightarrow-2\cdot\sqrt{x-2}=-1\)

\(\Leftrightarrow\sqrt{x-2}=\dfrac{1}{2}\)

\(\Leftrightarrow x-2=\dfrac{1}{4}\)

hay \(x=\dfrac{9}{4}\)

a) \(\left(x+1+\dfrac{1}{x}\right)^2=\left(x-1-\dfrac{1}{x}\right)^2\\ \Leftrightarrow\left(x+1+\dfrac{1}{x}\right)^2-\left(x-1-\dfrac{1}{x}\right)^2=0\\ \Leftrightarrow\left(x+1+\dfrac{1}{x}-x+1+\dfrac{1}{x}\right)\left(x+1+\dfrac{1}{x}+x-1-\dfrac{1}{x}\right)=0\\ \Leftrightarrow2\left(1+\dfrac{1}{x}\right)\cdot2x=0\\ \Leftrightarrow4x\left(1+\dfrac{1}{x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(S=\left\{-1;0\right\}\) là tập nghiệm của pt.

b) Ta có: \(\left(x-1\right)^2+3x^2=0\)

\(\Leftrightarrow x^2-2x+1+3x^2=0\)

\(\Leftrightarrow4x^2-2x+1=0\)

\(\text{Δ}=\left(-2\right)^2-4\cdot4\cdot1=4-16=-12< 0\)

=> Phương trình vô nghiệm

Vậy: \(S=\varnothing\)

a.

Kiểm tra lại đề bài, đề bài không đúng

b.

ĐKXĐ: \(x\ge0\)

\(1+3\sqrt{x}=4x+\sqrt{x+2}\)

\(\Rightarrow4x-1-\left(3\sqrt{x}-\sqrt{x+2}\right)=0\)

\(\Leftrightarrow4x-1-\dfrac{2\left(4x-1\right)}{3\sqrt{x}+\sqrt{x+2}}=0\)

\(\Leftrightarrow\left(4x-1\right)\left(1-\dfrac{2}{3\sqrt{x}+\sqrt{x+2}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\Rightarrow x...\\3\sqrt{x}+\sqrt{x+2}=2\left(1\right)\end{matrix}\right.\)

Xét (1): \(\Leftrightarrow10x+2+6\sqrt{x^2+2x}=4\)

\(\Leftrightarrow3\sqrt{x^2+2x}=1-5x\) (\(x\le\dfrac{1}{5}\))

\(\Leftrightarrow16x^2-28x+1=0\Rightarrow x=\dfrac{7-3\sqrt{5}}{8}\)

`a,x(x+3)-(2x-1).(x+30)=0`
`<=>x^2+3x-(2x^2+59x-30)=0`
`<=>x^2+56x-30=0`
`<=>x^2+56x+28^2=28^2+30`
`<=>(x+28)^2=28^2+30`
`<=>x=+-sqrt{28^2+30}-28`
`b,x(x-3)-5(x-3)=0`
`<=>(x-3)(x-5)=0`
`<=>` $\left[ \begin{array}{l}x=3\\x=5\end{array} \right.$
`c)1/(x-1)+5/(x-2)=(3x)/((x-1)(x-2))`
`đk:x ne 1,2`
`pt<=>x-2+5(x-1)=3x`
`<=>x-2+5x-5=3x`
`<=>6x-7=3x`
`<=>3x=7`
`<=>x=7/3`
`d)(x-1)/(x+1)+(x+1)/(x-1)=(4-2x^2)/(x^2-1)`
`đk:x ne +-1`
`pt<=>(x-1)^2+(x+1)^2=4-2x^2`
`<=>2x^2+2=4-2x^2`
`<=>4x^2=2`
`<=>x^2=1/2`
`<=>x=+-sqrt{1/2}`

a: =>-3x=-12

=>x=4

b: =>3(3x+2)-3x-1=12x+10

=>9x+6-3x-1=12x+10

=>12x+10=6x+5

=>6x=-5

=>x=-5/6

c: =>x(x+1)+x(x-3)=4x

=>x^2+x+x^2-3x-4x=0

=>2x^2-6x=0

=>2x(x-3)=0

=>x=3(loại) hoặc x=0(nhận)

a, \(\left|sinx+\dfrac{1}{2}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow sin^2x+sinx+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow sin^2x+sinx=0\)

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

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

b, \(tan^2\left(x+\dfrac{\pi}{6}\right)=3\)

\(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=\pm\sqrt{3}\)

\(\Leftrightarrow x+\dfrac{\pi}{6}=\pm\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

c, \(2sin\left(4x-\dfrac{\pi}{3}\right)-1=0\)

\(\Leftrightarrow sin\left(4x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\4x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{7\pi}{24}+\dfrac{k\pi}{2}\end{matrix}\right.\)

a: \(\left\{{}\begin{matrix}\dfrac{x}{35}-y=2\\y-\dfrac{x}{50}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x-35y}{35}=2\\\dfrac{50y-x}{50}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-35y=70\\-x+50y=50\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15y=120\\x-35y=70\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=8\\x=70+35y=70+35\cdot8=350\end{matrix}\right.\)

b: ĐKXĐ: x<>0 và y<>0

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=\dfrac{3}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{y}=\dfrac{3}{16}-\dfrac{1}{4}=\dfrac{-1}{16}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=48\\\dfrac{1}{x}=\dfrac{1}{16}-\dfrac{1}{48}=\dfrac{2}{48}=\dfrac{1}{24}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=24\\y=48\end{matrix}\right.\left(nhận\right)\)