Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\left(4x-7\right)\left(16x^2+28+49\right)=\left(4x\right)^3-7^3=64x^3-343\)\(b,\left(3x+1\right)\left(9x^2-3x+1\right)-9x\left(3x^2+9x\right)\)
\(=27x^3+1-27x^3+9x=9x+1\)
Ta có:
2x(2x-1)2-3x(x+3)(x-3)-4x(x+1)2
=2x(4x2-4x+1)-3x(x2-9)-4x(x2+2x+1)
=8x3-8x2+2x-3x3+27x-4x3-8x2+4x
=x3-16x2 +33x
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
\(A=\left(4x+5\right)^2-\left(3x-7\right)^2-\left(4x-1\right)\left(4x+1\right)\)
\(=\left(4x\right)^2+2.4x.5+5^2-\left[\left(3x\right)^2-2.3x.7+7^2\right]-\left[\left(4x\right)^2-1^2\right]\)
\(=16x^2+40x+25-\left(9x^2-42x+49\right)-\left(16x^2-1\right)\)
\(=16x^2+40x+25-9x^2+42x-49-16x^2+1=-9x^2+82x-23\)
A=(4x+5)2-(3x-7)2-(4x-1)(4x+1)
=16x2+40x+25-9x2+42x-49-16x2+1
=(16x2-9x2-16x2)+(40x+42x)+(25-49+1)
=-9x2+82x-23