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

\(a,=\dfrac{2\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x}\\ b,=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\\ =\dfrac{\left(1-3x\right)\left(2x-1\right)+\left(2x-1\right)\left(3x-2\right)}{2x\left(2x-1\right)}\\ =\dfrac{\left(2x-1\right)\left(1-3x+3x-2\right)}{2x}=\dfrac{-1}{2x}\)

a \(\dfrac{2x-4}{x^2-2x}\)= \(\dfrac{2\left(x-2\right)}{x\left(x-2\right)}\)=\(\dfrac{2}{x}\)

Lời giải:

$(2x-3)(x^2+1)=0$

\(\Leftrightarrow \left[\begin{matrix} 2x-3=0\\ x^2+1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{3}{2}(\text{chọn})\\ x^2=-1<0(\text{vô lý})\end{matrix}\right.\)

Vậy pt có nghiệm $x=\frac{3}{2}$

\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{2.\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{1.\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow2x-4-x+1=3x-11\\ \Leftrightarrow x-3=3x-11\\ \Leftrightarrow x-3x=-11+3\\ \Leftrightarrow-2x=-8\\ \Leftrightarrow x=4\)

Vậy tập nghiệm của phương trình là S = { 4 }

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

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

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

b: Để P=-1/2 thì 1/2(x+1)=-1/2

=>x+1=-1

=>x=-2

\(4x^2+4x+1+4x+2-2x^2-x\le0\)

\(\Leftrightarrow2x^2+7x+3\le0\Leftrightarrow\left(2x+1\right)\left(x+3\right)\le0\)

TH1 : \(\left\{{}\begin{matrix}2x+1\ge0\\x+3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\le-3\end{matrix}\right.\)<=> -1/2 =< x =< -3 

TH2 : \(\left\{{}\begin{matrix}2x+1\le0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\x\ge-3\end{matrix}\right.\)( vô lí ) 

\(y=cosx-3.sinx\le\sqrt{\left(1+\left(-3\right)^2\right)\left(cos^2x+sin^2x\right)}=\sqrt{10}\)

\(\Rightarrow y_{max}=\sqrt{10}\)

Giải nhanh giúp mình,please

\(\frac{x-4}{-5}=\frac{1-2x}{3}\)

Nhân cả 2 vế với 15 ,ta được:

\(\frac{15.\left(x-4\right)}{-5}=\frac{15.\left(1-2x\right)}{3}\)

\(\Leftrightarrow\left(-3\right).\left(x-4\right)=5.\left(1-2x\right)\)

\(\Leftrightarrow-3x+12=5-10x\)

\(\Leftrightarrow-3x+10x=5-12\)

\(\Leftrightarrow7x=-7\)

\(\Leftrightarrow x=-1\)

Vậy x=-1

(𝑥−4)/−5=(1−2𝑥)/3

−15⋅𝑥−4−5=−15⋅−2𝑥+13

−15⋅𝑥−4−5=−15⋅−2𝑥+13

3(𝑥−4)=−5(−2𝑥+1)

3(x-4)=-5(-2x+1)

3(𝑥−4)=−5(−2𝑥+1)

3𝑥−12=−5(−2𝑥+1)

3𝑥−12=−5(−2𝑥+1)

3𝑥−12=10𝑥−5

𝑥 = -1