rút gọn bt sau: (2x-3)2-(2x+3)2
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\(A=\frac{2x^2+4x}{x^3-4x}+\frac{x^2-4}{x^2+2x}+\frac{2}{2-x}\left(x\ne0;x\ne\pm2\right)\)
\(A=\frac{2x^2+4x}{x\left(x^2-4\right)}+\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}-\frac{2}{x-2}\)
\(A=\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}+\frac{x^3-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}-\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x^2+4x+x^3-2x^2-4x+8-2x^2-4x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x\left(x+2\right)+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x+8}{x\left(x-2\right)}\)
Vậy \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)
b) \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)
Ta có: x=4 (tmđk) thay vào A ta có:
\(A=\frac{-2\cdot4+8}{4\left(4-2\right)}=\frac{-8+8}{4\cdot2}=\frac{0}{8}=0\)
Vậy A=0 với x=4
A = (2x + 3)2 - (2x - 3)(2x + 3)
= 4x2 + 12x + 9 - 4x2 + 9
= 12x +18
Rút gọn biểu thức sau A=(2x+3)^2-(2x-3)×(2x+3)
trả lời:
\(A=\left(2x+3^2\right)-\left(2x-3\right).\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
\(=12x+8\)
\(=20\)
Vậy \(x=20\)
học tốt
( 2x - 3 )( 2x + 3 ) - ( 3x - 4 )2 + ( 2x + 3 )2
= 4x2 - 9 - ( 9x2 - 24x + 16 ) + 4x2 + 12x + 9
= 8x2 + 12x - 9x2 + 24x - 16
= -x2 + 36x - 16
`@` `\text {Ans}`
`\downarrow`
`(2x - 3)^2 - (2x + 3)^2`
`= 4x^2 - 12x + 9 - (4x^2 + 12x + 9)`
`= 4x^2 - 12x + 9 - 4x^2 - 12x - 9`
`= (4x^2 - 4x^2) + (-12x - 12x) + (9-9)`
`= -24x`
____
`@` CT:
`(A + B)^2 = A^2 + 2AB + B^2`
`(A - B)^2 = A^2 - 2AB + B^2`
\(\left(2x-3\right)^2-\left(2x+3\right)^2\)
\(=\left[\left(2x-3\right)+\left(2x+3\right)\right]\left[\left(2x-3\right)-\left(2x+3\right)\right]\)
\(=\left(2x-3+2x+3\right)\left(2x-3-2x-3\right)\)
\(=4x\cdot-6\)
\(=-24x\)