(2x+3).(10x+2)=(5x+2).(4x+5)
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2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
2x+3/5x+2 = 4x+5/10x+2
<=> (2x+3)(10x+2)=(5x+2)(4x+5)
<=>2x(10x+2)+3(10x+2)= 5x(4x+5)+2(4x+5)
<=> 20x^2+4x+20x+6 = 20x^2+25x+9x+10
<=> 20x^2+4x+20x+6 - (20x^2+25x+9x+10)=0 => 20x^2+24x+6-(20x^2+34x+10)=0
<=> -10x-4=0
<=>-10x=4
<=> x= -4/10
2x+3/5x+2=4x+5/10x+2
=> (2x+3)(10x+2)=(5x+2)(4x+5)
=> 20x^2+4x+30x+6=10x^2+25x+8x+10 ( Vì cả hai vế đều có 10x^2 nên ta xóa đi )
=> 34x+6=33x+10
=> 34x-33x=-6+10
=> x=4
\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(4x+5\right)\left(5x+2\right)\Leftrightarrow20x^2+34x+6=20x^2+33x+10\Leftrightarrow x=4\)
a: \(=8x^3-10x^5+10x^5-5x^3=3x^3\)
b: \(=x^3-2x^2+4x-2x^2+4x-8+x^2+2x-4x-8\)
\(=x^3-3x^2+4x-16\)
\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
=> (2x + 3)(10x + 2) = (5x + 2)(4x + 5)
=> 20x2 + 4x + 30x + 6 = 20x2 + 25x + 8x + 10
=> 34x + 6 = 33x + 10 (bớt 2 vế đi 20x2)
=> x = 4
\(\left(2x+3\right)\cdot\left(10x+2\right)=\left(5x+2\right)\cdot\left(4x+5\right)\)
\(\Leftrightarrow20x^2+4x+30x+6-20x^2-25x-8x-10=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
\(< =>20x^2+4x+30x+6=20x^2+25x+8x+10\)
\(< =>34x+6-25x-8x-10=0\)
\(< =>x-4=0< =>x=4\)