Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 3(x - 1)2 - 3x(x - 5) = 3(x2 - 2x + 1) - 3x2 + 15x = 3x2 - 6x + 3 - 3x2 + 15x = 9x + 3 = 21 => x = (21 - 3) : 9 = 18 : 9 = 2
b) 3(x + 2)2 + (2x - 1)2 - 7(x + 3)(x - 3) = 3(x2 + 4x + 4) + 4x2 - 4x + 1 - 7(x2 - 9) = 3x2 + 12x + 12 + 4x2 - 4x + 1 - 7x2 + 63
= 8x + 76 = 36 => x = (36 - 76) : 8 = -40 : 8 = -5
\(A=x^2-2x+1-x^2+4=5-2x\)
\(B=27x^3+8-x^2+9=27x^3-x^2+17\)
\(C=3x^2y-6xy^2-2x\left(x^2-2x^2y+x^2y^2\right)=3x^2y-6xy^2-2x^3+4x^3y-2x^3y^2\)
Em chỉ cần nhớ hằng đẳng thức và áp dụng là biến đổi được ^^
a)
(x3 – 3x2+3x-1) – (x3+9) +(3x2-12) = 2
3x-22 = 2
3x =24
=> x= 8
b)
x3+6x2+12x+8-6x2-x3-x2+x2 = 5
12x+8 = 5
12x = -3
=>x = -1/4
a) ( x - 1 )3 - ( x + 3 )(x2 - 3x + 9 ) + 3( x2 - 4 ) = 2
(x3-3x2\(\times\)1+3x\(\times\)12-13)-(x3+33)+(3x2-3\(\times\)4)=2
x3-3x2+3x-1-x3-9+3x2-12=2
3x-40=2
3x=42
x=14
b ) ( x + 2 )3 - 6x2 - x2 ( x + 1 ) + x2 = 5
(x3+3\(\times\)2x2+3x\(\times\)22+23)-6x2-(x3+x2)+x2=5
x3+6x2+12x+8-6x2-x3-x2+x2=5
12x+8=5
12x=\(-\)3
x=\(-\frac{1}{4}\)
bài dễ mà 
\(a,\left(x+3\right).\left(x^2-3x+9\right)-\left(54+x^3\right)=x^3+27-54-x^3=-27.\)
\(b,8x^3+y^3-8x^3+y^3=2y^3\)
\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
a) (a+b)3- (a-b)3- 2ab
=a3+3a2b+3ab2+b3-(a3-3a2b+3ab2-b3)-2ab
=a3+3a2b+3ab2+b3-a3+3a2b-3ab2+b3-2ab
=2b3+6a2b-2ab
b) (x-2). (x2+2x+4) - x.(x2-1)+x+5
=x3-8-x3+x+x+5
=2x-3
Bài 1:
a) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)
\(\Rightarrow x^3-3x^2+3x-1+2^3-x^3+3x^2+6x=17\)
\(\Rightarrow9x+7=17\)
\(\Rightarrow9x=17-7=10\)
\(\Rightarrow x=\dfrac{10}{9}\)
b) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Rightarrow x^3+2^3-x^3+2x=15\)
\(\Rightarrow8+2x=15\)
\(\Rightarrow2x=15-8=7\)
\(\Rightarrow x=\dfrac{7}{2}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Rightarrow x^3-3x^2.3+3x.3^2-3^3-x^3+3^3+9\left(x^2+2x+1\right)=15\)
\(\Rightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Rightarrow45x+9=15\)
\(\Rightarrow45x=6\)
\(\Rightarrow x=\dfrac{6}{45}=\dfrac{2}{15}\)
d) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Rightarrow x\left(x^2-5^2\right)-x^3-2^3=3\)
\(\Rightarrow x^3-25x-x^3-8=3\)
\(\Rightarrow-25x-8=3\)
\(\Rightarrow-25x=3+8=11\)
\(\Rightarrow x=-\dfrac{11}{25}\)
Bài 2:
a) Ta có:
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\)
\(B=2^{16}-1\)
Vì 216 - 1 < 216
=> B < A
b) Ta có:
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{128}-1\right)\)
Vì 1/2( 3128 - 1) < 3128 - 1
=> A < B
\(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=x^3-4x^2+16x+4x^2-16x+64\)
\(=x^3+64\)
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^2+3x^2y+9xy^2-3x^2y-9xy^2-27y^3\)
\(=\)\(x^2-27y^3\)
\(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3xy}+4y^2\right)\)
\(=\)\(\frac{x^3}{27}-\frac{2}{9xy}+\frac{4xy^2}{3}+\frac{2x^2y}{9}-\frac{4y}{3xy}+8y^3\)
làm nốt nha
\(\left(5-x^2\right)\left(5+x^2\right)-x^2\left(2-x^2\right)\)
\(=25-x^4-2x^2+x^4\)
\(=25-2x^2\)
Câu này áp dụng hằng đẳng thức quái gì, sửa lại
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(1+x\right)\)
\(=x^3+27-x\left(x^2-1\right)\)
\(=x^3+27-x^3+x=27+x\)
a) số lẻ wa
b)(x - 1)3 - (x + 3) . (x2 - 3x +9) + 3 . (x + 2) . (x - 2) = 2
\(VT=3x-40\)
\(\Leftrightarrow3x-40=2\)
\(\Leftrightarrow3x=42\)
\(\Leftrightarrow x=14\)
Ai giúp mình câu a với !!!