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1: (3x+2)(x+2)(2x-1)
=(3x^2+6x+2x+4)(2x-1)
=(3x^2+8x+4)(2x-1)
=6x^3-3x^2+16x^2-8x+8x-4
=6x^3+13x^2-4
2: (5x+1)(x-1)+3x(2x+2)
=5x^2-5x+x-1+6x^2+6x
=11x^2+10x-1
3: 4x(2x+1)(x-1)+(x+5)(x-3)
=4x(2x^2-2x+x-1)+x^2+2x-15
=8x^3-4x^2-4x+x^2+2x-15
=8x^3-3x^2-2x-15
4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)
=(2x-1)(x^2-4)+3x^2-4x+1
=2x^3-8x-x^2+4+3x^2-4x+1
=2x^3+2x^2-12x+5
a) \(({x^2} + 2x + 3) + (3{x^2} - 5x + 1) = ({x^2} + 3{x^2}) + (2x - 5x) + (3 + 1) = 4{x^2} - 3x + 4\);
b) \(\begin{array}{l}(4{x^3} - 2{x^2} - 6) - ({x^3} - 7{x^2} + x - 5) = 4{x^3} - 2{x^2} - 6 - {x^3} + 7{x^2} - x + 5\\ = (4{x^3} - {x^3}) + ( - 2{x^2} + 7{x^2}) - x + ( - 6 + 5) = 3{x^3} + 5{x^2} - x - 1\end{array}\);
c) \(\begin{array}{l} - 3{x^2}(6{x^2} - 8x + 1) = - 3{x^2}.6{x^2} - - 3{x^2}.8x + - 3{x^2}.1\\ = - 18{x^{2 + 2}} + 24{x^{2 + 1}} - 3{x^2} = - 18{x^4} + 24{x^3} - 3{x^2}\end{array}\);
d) \(\begin{array}{l}(4{x^2} + 2x + 1)(2x - 1) = (4{x^2} + 2x + 1).2x - (4{x^2} + 2x + 1).1 = 4{x^2}.2x + 2x.2x + 1.2x - 4{x^2} - 2x - 1\\ = 8{x^{2 + 1}} + 4{x^{1 + 1}} + 2x - 4{x^2} - 2x - 1 = 8{x^3} + 4{x^2} + 2x - 4{x^2} - 2x - 1 = 8{x^3} - 1\end{array}\);
e) \(\begin{array}{l}({x^6} - 2{x^4} + {x^2}):( - 2{x^2}) = {x^6}:( - 2{x^2}) - 2{x^4}:( - 2{x^2}) + {x^2}:( - 2{x^2})\\ = - \dfrac{1}{2}{x^{6 - 2}} + {x^{4 - 2}} - \dfrac{1}{2}{x^{2 - 2}} = - \dfrac{1}{2}{x^4} + {x^2} - \dfrac{1}{2}.\end{array}\);
g)
\(({x^5} - {x^4} - 2{x^3}):({x^2} + x)=x^3-2x^2\)
1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow-2x\left(x+1\right)=0\)
Vì -2≠0
nên \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: x∈{0;-1}
2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)
\(\Leftrightarrow11x-2=0\)
\(\Leftrightarrow11x=2\)
hay \(x=\frac{2}{11}\)
Vậy: \(x=\frac{2}{11}\)
3) Ta có: \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0\)
\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}\)
hay \(x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}\)
Vậy: \(x=\frac{-13}{4}\)
4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)
1. \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x=3x^2-x\)
\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)
\(\Leftrightarrow-2x-2x^2=0\)
\(\Leftrightarrow-2x\left(1+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
2. \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x=12-x\)
\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)
\(\Leftrightarrow-2+11x=0\)
\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)
3. \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0\)
\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}\)
4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Rightarrow x-\frac{4}{5}=-\frac{3}{5}\) hoặc \(x-\frac{4}{5}=\frac{3}{5}\)
\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)
\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1
=>-16x-34=x-1
=>-17x=33
=>x=-33/17
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
e: =>8x+16-5x^2-10x+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6
=>4x^2+16x-20-4x^2-10x+4=0
=>6x=16
=>x=8/3
a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)
\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)
\(=2x^2+x+1\)
b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)
c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)
\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)
d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)
\(=x^2-2x-5\)
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
