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1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
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`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
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`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
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`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
1, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)
\(\Leftrightarrow-32x^2+8x+4x^3-13=0\)( vô nghiệm )
2, \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
\(\Leftrightarrow12x^3-7x^2-10x-7x^2-35x=-2x^2+11x-12+12x^3+2x^2\)
\(\Leftrightarrow12x^3-14x^2-45x=11x-12+12x^3\)
\(\Leftrightarrow-14x^2-56x-12=0\)( vô nghiệm )
Mình làm riêng ra nhá , chứ nhiều quá nên thông cảm cho mình :))
1. \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
=> \(-4x^2+28x+4x^3-20x=28x^2-13\)
=> \(-4x^2+4x^3+\left(28x-20x\right)=28x^2-13\)
=> \(-4x^2+4x^3+8x-28x^2+13=0\)
=> \(\left(-4x^2-28x^2\right)+4x^3+8x+13=0\)
=> \(-32x^2+4x^3+8x+13=0\)
=> vô nghiệm
2. \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
=> \(4x^2\left(3x+2\right)-5x\left(3x+2\right)-7x\left(x+5\right)=-4\left(-2x+3\right)+x\left(-2x+3\right)+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x=8x-12-2x^2+3x+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x-8x+12+2x^2-3x-12x^3-2x^2=0\)
=> \(\left(12x^3-12x^3\right)+\left(8x^2-15x^2-7x^2+2x^2-2x^2\right)+\left(-10x-35x-8x-3x\right)+12=0\)
=> \(-14x^2-56x+12=0\)
=> .... tự tìm
Câu c dấu bằng chỗ nào ?
\(a,\left(3x+4\right)\left(3x-4\right)-\left(2x+5\right)^2=\left(x-5\right)^2+\left(2x+1\right)^2-\left(x^2-2x\right)+\left(x-1\right)^2\\ \Leftrightarrow\left(9x^2-16\right)-\left(4x^2+20x+25\right)=x^2-10x+25+4x^2+4x+1-x^2+2x+x^2-2x+1\\ \Leftrightarrow9x^2-16-4x^2-20x-25=5x^2-6x+27\\ \Leftrightarrow5x^2-20x-41=5x^2-5x+27\\ \Leftrightarrow-15x=68\\ \Leftrightarrow x=-\dfrac{68}{15}\)Vậy..
Câu sau cũng tương tự nhé
a: \(\Leftrightarrow8x+16-5x^2-10x+4x^2-4x-8+2\left(x^2-4\right)=0\)
\(\Leftrightarrow-x^2-6x+8+2x^2-8=0\)
=>x^2-6x=0
=>x(x-6)=0
=>x=6 hoặc x=0
b: \(\Leftrightarrow24x^2+7x-6-4x^2-23x-28=10x^2+3x-1-33\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+34=0\)
=>\(10x^2-19x=0\)
=>x(10x-19)=0
=>x=0 hoặc x=19/10
a) \(\left(3x-2\right)^2-\left(3x-5\right)\left(3x+2\right)=11\)
\(\Leftrightarrow\left(9x^2-12x+4\right)-\left(9x^2+6x-15x-10\right)=11\)
\(\Leftrightarrow9x^2-12x+4-9x^2-6x+15x+10=11\)
\(\Leftrightarrow-3x+3=0\)
\(\Leftrightarrow-3x=-3\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
b) \(\left(4x-3\right)^2-\left(4x-5\right)\left(4x+5\right)=32\)
\(\Leftrightarrow\left(16x^2-24x+9\right)-\left(16x^2-25\right)=32\)
\(\Leftrightarrow16x^2-24x+9-16x^2+25=32\)
\(\Leftrightarrow-24x+2=0\)
\(\Leftrightarrow-24x=-2\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy \(S=\left\{\dfrac{1}{12}\right\}\)
c) \(\left(5x-2\right)^2-\left(5x+3\right)\left(5x-5\right)=1\)
\(\Leftrightarrow\left(25x^2-20x+4\right)-\left(25x^2-25x+15x-15\right)=1\)
\(\Leftrightarrow25x^2-20x+4-25x^2+25x-15x+15=1\)
\(\Leftrightarrow-10x+18=0\)
\(\Leftrightarrow-10x=-18\)
\(\Leftrightarrow x=\dfrac{9}{5}\)
Vậy \(S=\left\{\dfrac{9}{5}\right\}\)
d) \(\left(x-4\right)^2-\left(x-7\right)\left(2x-3\right)=5-x^2\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(2x^2-3x-14x+21\right)=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21-5+x^2=0\)
\(\Leftrightarrow9x-10=0\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\dfrac{10}{9}\)
Vậy \(S=\left\{\dfrac{10}{9}\right\}\)
Cho mk hỏi vs ! Câu a bn rút gọn hay bn lm kiểu j mak tự nhiên 11 lại lôi đâu ra số 0 vậy ? Gt hộ mk vs, mk vẫn chưa hiểu cách bn lm ở câu a cho lắm !
\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)
\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)
a: =>(x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>x=1 hoặc x=-5
b: =>(x-1)^2=25
=>x-1=5 hoặc x-1=-5
=>x=-4 hoặc x=6
c: =>25x^2+10x+1-25x^2+9=30
=>10x+10=30
=>x+1=3
=>x=2
d: =>x^3-1-x(x^2-4)=5
=>x^3-1-x^3+4x=5
=>4x=6
=>x=3/2
d: Ta có: \(4x\left(2x+3\right)-8x\left(x+4\right)\)
\(=8x^2+12x-8x^2-32x\)
=-20x
e: Ta có: \(2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)
\(=10x^2+4x+6x^2-2x-9x+3\)
\(=16x^2-7x+3\)
f: Ta có: \(x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3+4x^2+4x-x^3-3x^2-3x-1+3x^2-3\)
\(=4x^2+x-4\)
\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
\(25x^2-10x+1-25x^2+16=7\)
\(17-10x=7\)
\(10x=10\)
\(x=1\)