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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 \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)
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
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 ............
b) \(\dfrac{x-5}{2017}-1+\dfrac{x-2}{2020}-1=\dfrac{x-6}{2016}-1+\dfrac{x-68}{1954}-1\)
\(\dfrac{x-2022}{2017}+\dfrac{x-2002}{2020}=\dfrac{x-2022}{2016}+\dfrac{x-2022}{1954}\)
\(\Leftrightarrow\left(x-2022\right)\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\right)=0\)
\(\Leftrightarrow x-2022=0\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\ne0\right)\)
\(\Leftrightarrow x=2022\)
a)\(\frac{2x-5}{x+5}\)=3 ĐKXĐ: x khác -5
=> 2x-5=3(x+5)
<=>2x-5=3x+15
<=>-x=20
<=>x =-20
b)\(\frac{x2-6}{x}\)=x+\(\frac{3}{2}\)ĐKXĐ\(x\ne0\)
=>2(x2-6)=2x2+3x
<=>2x2-12=2x2+3x
<=>-3x=12
<=>x=-4
Để \(\frac{2x\left(3x-5\right)}{x^2+1}< 0\)
ta thấy x2+1 luôn dương với mọi x
nên 2x(3x-5) <0
TH1: \(\orbr{\begin{cases}2x< 0\\3x-5>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 0\\3x>5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x< 0\\x>\frac{5}{3}\end{cases}\left(ktm\right)}}\)
TH2: \(\orbr{\begin{cases}2x>0\\3x-5< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x>0\\3x< 5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x>0\\x< \frac{5}{3}\end{cases}\left(tm\right)}}\)
vậy \(0< x< \frac{5}{3}\)
THẤY ĐÚNG CHO MK 1 NẾU KO HIỂU THÌ ib NHA
\(\frac{2x\left(3x-5\right)}{x^2+1}< 0\)
\(\Rightarrow2x\left(3x-5\right)< 0\) ( vì \(x^2+1>0\))
\(\Rightarrow\hept{\begin{cases}2x< 0\\3x-5>0\end{cases}}\) hoặc \(\hept{\begin{cases}2x>0\\3x-5< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< 0\\x>\frac{5}{3}\end{cases}}\) hoặc \(\hept{\begin{cases}x>0\\x< \frac{5}{3}\end{cases}}\)
\(\Rightarrow0< x< \frac{5}{3}\)