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giúp e vớiiii
6 tháng 9 2021
a: ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+8x^3+1-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2-54=0\)
\(\Leftrightarrow9x^3+6x^2+27x-26-9x^3-6x^2-x=0\)
\(\Leftrightarrow26x=26\)
hay x=1
b: Ta có: \(\left(x-3\right)^3-\left(x^3-27\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x^2+2x+1\right)+3x^2+33=0\)
\(\Leftrightarrow-6x^2+27x+33+6x^2+12x+6=0\)
\(\Leftrightarrow39x=-39\)
hay x=-1
4 tháng 8 2016
Ta đặt: a/b = a/d =k
=> a = b.k, c=d.k
Ta có: a2 + a.c/c2 - a.c=b2 + b.d/d2 - b.d
Vế trái: => (b.k)2 + (b.k)(d.k)/(d.k)2 - (b.k)(d.k)
=> b2.k2 + k(b.d)/d2.k2 - k.(b.d)
Ta lược bỏ các chữ giống nhau, ta được:
=> b2/d2
Vế phải: b2 +b.d/d2 - b.d
Ta cũng lược bỏ những chữa giống nhau ta được:
=> b2/d2
Vậy a2 +a.c/c2 + a.c = b2 + b.d/d2 - b.d
5: \(\left(2x-3\right)^2-16\)
\(=\left(2x-3\right)^2-4^2\)
\(=\left(2x-3-4\right)\left(2x-3+4\right)\)
\(=\left(2x-7\right)\left(2x+1\right)\)
6: \(\left(2x+1\right)^2-\left(\dfrac{1}{2}x-7\right)^2\)
\(=\left(2x+1-\dfrac{1}{2}x+7\right)\left(2x+1+\dfrac{1}{2}x-7\right)\)
\(=\left(\dfrac{3}{2}x+8\right)\left(\dfrac{5}{2}x-6\right)\)
7: \(\left(x+7\right)^2-\left(4x+5\right)^2\)
\(=\left(x+7-4x-5\right)\left(x+7+4x+5\right)\)
\(=\left(-3x+2\right)\left(5x+12\right)\)
8: \(\left(\dfrac{1}{4}x+3\right)^2-\left(4x-\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{4}x+3-4x+\dfrac{1}{4}\right)\left(\dfrac{1}{4}x+3+4x-\dfrac{1}{4}\right)\)
\(=\left(-\dfrac{15}{4}x+\dfrac{13}{4}\right)\left(\dfrac{17}{4}x+\dfrac{11}{4}\right)\)
9: \(\left(\dfrac{1}{3}x+3\right)^2-\left(4x-\dfrac{2}{3}\right)^2\)
\(=\left(\dfrac{1}{3}x+3-4x+\dfrac{2}{3}\right)\left(\dfrac{1}{3}x+3+4x-\dfrac{2}{3}\right)\)
\(=\left(-\dfrac{11}{3}x+\dfrac{11}{3}\right)\left(\dfrac{13}{3}x+\dfrac{7}{3}\right)\)
\(=-\dfrac{11}{3}\left(x-1\right)\left(\dfrac{13}{3}x+\dfrac{7}{3}\right)\)
10: \(9\left(x+5\right)^2-4\left(2x+8\right)^2\)
\(=\left[3\left(x+5\right)\right]^2-\left[2\left(2x+8\right)\right]^2\)
\(=\left(3x+15\right)^2-\left(4x+16\right)^2\)
\(=\left(3x+15-4x-16\right)\left(3x+15+4x+16\right)\)
\(=\left(-x-1\right)\left(7x+31\right)\)
11: \(\left(x+5\right)^2-25\left(2x+8\right)^2\)
\(=\left(x+5\right)^2-\left[5\left(2x+8\right)\right]^2\)
\(=\left(x+5\right)^2-\left(10x+40\right)^2\)
\(=\left(x+5-10x-40\right)\left(x+5+10x+40\right)\)
\(=\left(-9x-35\right)\left(11x+45\right)\)
12: \(16\left(2x+5\right)^2-9\left(2x-1\right)^2\)
\(=\left[4\left(2x+5\right)\right]^2-\left[3\left(2x-1\right)\right]^2\)
\(=\left(8x+20\right)^2-\left(6x-3\right)^2\)
\(=\left(8x+20+6x-3\right)\left(8x+20-6x+3\right)\)
\(=\left(14x+17\right)\left(2x+23\right)\)
1: \(=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x\cdot2y=4xy\)
2: \(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)
3: \(=x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\)
\(=2x^3+6xy^2=2x\left(x^2+3y^2\right)\)
4: \(=\left(7-4x+5\right)\left(7+4x-5\right)=\left(12-4x\right)\left(4x+2\right)\)
\(=8\left(3-x\right)\left(2x+1\right)\)