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a: \(\Leftrightarrow\dfrac{3}{x-2}=\dfrac{2x-1}{x-2}-\dfrac{x\left(x-2\right)}{x-2}\)
=>3=2x-1-x^2+2x
=>3=-x^2+4x-1
=>x^2-4x+1+3=0
=>x^2-4x+4=0
=>x=2(loại)
b: =>(x+2)(2x-4)=x(2x+3)
=>2x^2-4x+4x-8=2x^2+3x
=>3x=-8
=>x=-8/3(nhận)
I,
\(B=\left(3x-1\right)^2-\left(x+7\right)^2-2\left(2x-5\right)\left(2x+5\right)\\ =9x^2-6x+1-x^2-14x-49-2\left(4x^2-25\right)\\ =8x^2-20x-48-8x^2+50\\ =-20x+2\)
II,
\(a,2x^2+6xy-10.Thayx=-4,y=3,tacó: 2\cdot\left(-4\right)^2+6\cdot\left(-4\right)\cdot3-10=-50\)
\(b,x\left(x+y\right)+y\left(x+y\right)=\left(x+y\right)\left(x+y\right)=\left(x+y\right)^2\\ Thayx=19,6;y=0,4tacó:\\ \left(19,6+0,4\right)^2=400\)
\(c,x\left(x-3\right)-y\left(3-x\right)=x\left(x-3\right)+y\left(x-3\right)=\left(x+y\right)\left(x-3\right)\\ Thayx=\frac{1}{3};y=\frac{8}{3},tacó:\\ \left(\frac{1}{3}+\frac{8}{3}\right)\left(\frac{1}{3}-3\right)=-8\)
\(d,2x^2\left(x^2+y^2\right)+2y^2\left(x^2+y^2\right)+5\left(x^2+y^2\right)\\ =\left(x^2+y^2\right)\left[2\left(x^2+y^2\right)+5\right]\\ Thayx^2+y^2=1,tacó:\\ 1\cdot\left(2\cdot1+5\right)=7\)
Bài 1:
\(\frac{15ab+5b^2}{9a^2-b^2}=\frac{5b\left(3a+b\right)}{\left(3a\right)^2-b^2}=\frac{5b\left(3a+b\right)}{\left(3a-b\right)\left(3a+b\right)}=\frac{5b}{3a-b}\)
\(\frac{3x^2-3y^2}{9x+9y}=\frac{3\left(x^2-y^2\right)}{9\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{3\left(x+y\right)}=\frac{x-y}{3}\)
\(\frac{m^2-4m+4}{2x-4}=\frac{\left(x-2\right)^2}{2\left(x-2\right)}=\frac{x-2}{2}\)
Giải các phương trình và bất phương trình sau :
1.1
a) \(2x+3=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
Tập nghiệm của pt là : \(S=\left\{\dfrac{-3}{2}\right\}\)
b) \(5x-3< 2x+9\)
\(\Leftrightarrow5x-2x< 3+9\)
\(\Leftrightarrow3x< 12\)
\(\Leftrightarrow x< 4\)
Tập nghiệm của BPT là : \(S=\left\{x|x< 4\right\}\)
1.2
a) \(3x+2=0\)
\(\Leftrightarrow3x=-2\)
\(\Leftrightarrow x=\dfrac{-2}{3}\)
Tập nghiệm của pt là : \(S=\left\{\dfrac{-2}{3}\right\}\)
b) \(-x+5>6-2x\)
\(\Leftrightarrow-x+2x>-5+6\)
\(\Leftrightarrow x>1\)
Tập nghiệm của BPT là : \(S=\left\{x|x>1\right\}\)
c) \(\dfrac{2x-5}{x+3}=4\)
ĐKXĐ : \(x+3\ne0\Rightarrow x\ne-3\)
\(\Leftrightarrow\dfrac{2x-5}{x+3}=\dfrac{4\left(x+3\right)}{x+3}\)
\(\Rightarrow2x-5=4x+12\)
\(\Leftrightarrow2x-4x=5+12\)
\(\Leftrightarrow-2x=17\)
\(\Leftrightarrow x=\dfrac{-17}{2}\)
Tập nghiệm của pt là : \(S=\left\{\dfrac{-17}{2}\right\}\)
d) \(\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Tập nghiệm của pt là : \(S=\left\{-2;3\right\}\)
1.3
a)\(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x+5-x-2\right).\left(2x+5+x+2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\3x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{-7}{3}\end{matrix}\right.\)
Tập nghiệm của pt là : \(S=\left\{\dfrac{-7}{3};-3\right\}\)
b) \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow\left(x^2-3x\right)-\left(2x-6\right)=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Tập nghiệm của pt là : \(S=\left\{2;3\right\}\)
\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)
\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)
\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Câu 1:
$(2x^2-3)(x+5)=2x^2(x+5)-3(x+5)=2x^3+10x^2-3x-15$
Câu 2:
a.
$(x+3)^2=x^2+2.x.3+3^2=x^2+6x+9$
b.
$y^2-25=y^2-25$
a: =>3(3x-7)+2(x+1)=-96
=>9x-21+2x+2=-96
=>11x-19=-96
=>11x=-96+19=-75
=>x=-75/11
b: \(x-\dfrac{x+1}{3}=\dfrac{2x+1}{5}\)
=>15x-5(x+1)=3(2x+1)
=>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
a)
\(\dfrac{3x-7}{2}+\dfrac{x+1}{3}=-16\)
\(< =>9x-21+2x+2=-96\)
\(< =>9x+2x=-96+21-2\\ < =>11x=-77\\ < =>x=-7\)
b)
\(\dfrac{x-x+1}{3}=\dfrac{2x+1}{5}\\ < =>5=6x+3\\ < =>6x=5-3\\ < =>6x=2\\ < =>x=\dfrac{1}{3}\)
a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)
\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)
=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)
=>30y+25=25y
=>5y=-25
=>y=-5(loại)
b: \(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
=>x^2+x+x^2-3x-4x=0
=>2x^2-6x=0
=>2x(x-3)=0
=>x=0(nhận) hoặc x=3(loại)
c: =>x^2-9-6(2x+7)=-13(x+3)
=>x^2-9-12x-42+13x+39=0
=>x^2+x-6=0
=>(x+3)(x-2)=0
=>x=2(nhận) hoặc x=-3(loại)