Bai1:

\(-2x+\frac{3}{5}\le\frac{3\left(2x-7\right)}{3}\Leftrightarrow-10x+3\le5\left(2x-7\right)\Leftrightarrow-10x+3\le10x-35\)

\(\Leftrightarrow\left(10+10\right)x\ge3+35\Rightarrow x\ge\frac{38}{20}=\frac{19}{10}\)

Bài

\(\left\{\begin{matrix}x+m-1>0\\3m-2-x>0\end{matrix}\right.\Leftrightarrow\left(I\right)\left\{\begin{matrix}x>1-m\\x< 3m-2\end{matrix}\right.\)

Hệ (I) có nghiệm cần m thỏa mãn:

\(1-m< 3m-2\Leftrightarrow1+2< 3m+m\Rightarrow m>\frac{3}{2}\)

Kết luận: để hệ có nghiệm cần: m>3/2

a) Ta có: 3x-6=0

⇔3(x-2)=0

mà 3≠0

nên x-2=0

hay x=2

Vậy: x=2

b) Ta có: (2x+6)(2x+12)=0

⇔\(2\left(x+3\right)\cdot2\cdot\left(x+6\right)=0\)

mà 2≠0

nên \(\left[{}\begin{matrix}x+3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-6\end{matrix}\right.\)

Vậy: x∈{-3;-6}

c) Ta có: 2x-36=0

⇔2(x-18)=0

mà 2≠0

nên x-18=0

hay x=18

Vậy: x=18

d) ĐKXĐ: x∉{-1;2}

Ta có: \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{-15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{-15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow x-2-5\left(x+1\right)=-15\)

\(\Leftrightarrow x-2-5x-5+15=0\)

\(\Leftrightarrow-4x+8=0\)

\(\Leftrightarrow-4\left(x-2\right)=0\)

mà -4≠0

nên x-2=0

hay x=2(ktm)

Vậy: x∈∅

Bài 1:

a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)

\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)

\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)

Suy ra: \(12x-45-12x^2+45x=0\)

\(\Leftrightarrow-12x^2+57x-45=0\)

\(\Leftrightarrow-12x^2+12x+45x-45=0\)

\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)

\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)

mà \(-3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)

b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

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

\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)

Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)

\(\Leftrightarrow x^2-13x-3x+39=0\)

\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)

Vậy: Tập nghiệm S={3;13}

c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)

\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)

\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)

Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)

\(\Leftrightarrow-21x^2+26x+11=0\)

\(\Leftrightarrow-21x^2-7x+33x+11=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)

1.

\(\Leftrightarrow\sqrt{3}cos^2x-\sqrt{3}+cos^2x+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx=0\)

\(\Leftrightarrow-\sqrt{3}sin^2x+cosx+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+\sqrt{3}sinx\right)-\left(cosx-sinx\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+\sqrt{3}sinx-1\right)=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx-\frac{1}{2}\right)=0\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\left[sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)

2/

\(\Leftrightarrow3sinx-4sin^3x-\sqrt{3}cosx=2sinx\)

\(\Leftrightarrow4sin^3x-sinx+\sqrt{3}cosx=0\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow4tan^3x-tanx\left(1+tan^2x\right)+\sqrt{3}\left(1+tan^2x\right)=0\)

\(\Leftrightarrow3tan^3x+\sqrt{3}tan^2x-tanx+\sqrt{3}=0\)

Bạn xem lại đề, pt bậc 3 này ko giải được (nghiệm rất xấu)

a: =(x-3)(2x+5)

b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)

=>(x-2)(5-x)=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1

sorry b, phải là cái này nha

a, Ta có: \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)

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

\(Đkxđ:\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)

\(Pt\Leftrightarrow x\left(x+2\right)-2=x-2\)

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

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

Vậy .........

\(b,Đkxđ:x\ne-5\)

Ta có: \(\frac{2x-5}{x+5}=3\)

\(\Leftrightarrow2x-5=3\left(x+5\right)\)

\(\Leftrightarrow x=20\left(tmđk\right)\)

Vậy .........

c, \(Đkxđ:x\ne3\)

Ta có: \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)

\(\Leftrightarrow x^2+2x-3x-6=0\)

\(\Leftrightarrow x^2-x-6=0\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=3\left(ktmđk\right)\end{matrix}\right.\)

Vậy ............