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a) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
= \(\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
= \(x^4+4x^2+4-x^4+16\)
= \(4x^2+20\)
b) \(\left(x+2y\right)^2-\left(x-2y\right)^2\)
= \(\left(x+2y-x+2y\right)\left(x+2y+x-2y\right)\)
= \(4y\cdot2x=8xy\)
a/ \(\left(x-2y\right)^2+3\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(x-2y+3x-6y\right)=\left(x-2y\right)\left(4x-8y\right)\)
\(=4\left(x-2y\right)\left(x-2y\right)=4\left(x-2y\right)^2\)
b/ \(\left(y^2+1\right)\left(y+2\right)-\left(y+2\right)\left(y^2-2y+4\right)\)
\(=y^3+2y^2+y+2-y^3-8\)
\(=2y^2+y-6=2y^2+4y-3y-6\)
\(=\left(y+2\right)\left(2y-3\right)\)
riêng câu b mình có sửa đề lại, bn xem có đúng hong nha. Chúc bn hc tốt nhé ^^
a) A = (x + 2y)(x^2 - 2xy + 4y^2) - 8(x^3 + y^3)
A = x(x^2 - 2xy + 4y^2) + 2y(x^2 - 2xy + 4y^2) - 8(x^3 + y^3)
A = x^3 - 2x^2y + 4xy^2 + 2x^2y - 4xy^2 + 8y^3 - 8x^3 - 8y^3
A = -7x^3
b) B = (2x + y)^3 - (8x^3 + y^3) - 2x^2y
B = (2x + y)[(2x)^2 + 2.2xy + y^2] - 8x^3 - y^3 - 2x^2y
B = 2x[(2x)^2 + 2.2xy + y^2] + y[(2x)^2 + 2.2xy + y^3] - 8x^3 - y^3 - 2x^2y
B = 8x^3 + 8x^2y + 2xy^2 + 4x^2y + y^3 - 8x^3 - y^3 - 2x^2y
B = 10x^2y + 6xy^2
Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)
Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm
a) \(A=x^2-8x+17=\left(x-4\right)^2+1\ge1\)
Vậy MIN A = 1 khi x = 4
b) \(T=x^2-4x+7=\left(x-2\right)^2+3\ge3\)
Vậy MIN T = 3 khi x = 2
c) \(H=3x^2+6x-1=3\left(x+1\right)^2-4\ge-4\)
Vậy MIN H = -4 khi x = -1
d) \(E=x^2+y^2-4\left(x+y\right)+16=\left(x-2\right)^2+\left(y-2\right)^2+8\ge8\)
Vậy MIN E = 8 khi x = y = 2
e) \(K=4x^2+y^2-4x-2y+3=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)
Vậy MIN K = 1 khi x = 1/2; y = 1
f) \(M=\frac{3}{2}x^2+x+1=\frac{3}{2}\left(x+\frac{1}{3}\right)^2+\frac{5}{6}\ge\frac{5}{6}\)
Vậy MIN M = 5/6 khi x = -1/3
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Bài 1 :
a) (3a+4b)3+(3a-4b)3-48a2b2
=27a3+108a2b+144ab2+64b3+27a3-108a2b+144ab2-64b3-48a2b2
=54a3+288ab2-48a2b2
=2a(27a2+144b2-24ab)
b) (5x+2y)(5x-2y)+(2x-y)3+(2x+y)3
=25x2-4y2+8x3-12x2y+6xy2-y3+8x3+12x2y+6xy2+y3
=16x3+25x2-y2+12xy2
=x2(16x+25)-y2(1-12x)
Bài 2 :
\(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
b)\(x^3-4x^2+3x=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{3}\\x=1\end{cases}}\)
c)Nếu đề đổi thành =1 thì có vẻ hợp lí hơn
d)\(\left(3x-1\right)^3-3\left(3x+2\right)^2+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-3\left(9x^2+12x+4\right)+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-27x^2-36x-12+13=0\)
\(\Leftrightarrow27x^3-54x^2-27x=0\)
\(\Leftrightarrow27x\left(x^2-2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}27x=0\\x^2-2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\-\left(x^2+2x+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\-\left(x+1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
#H
Rút gọn biểu thức:
a/ (x2+2)2 - (x+2)(x-2)(x2+4)
= (x2+2)2- (x2-22)(x2+4)
= (x2+2)2- (x2-4)(x2+4)
=(x2+2)2- (x4- 42)
= x4+4x2+4-x4+16
= 4x2+20
=4(x2+5)
b/ (x+2y)2-(x-2y)2
= x2+4xy+4y2-x2+4xy-y2
= 8xy
A. \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^4+4x^2+4\right)-\left[\left(x^2-4\right)\left(x^2+4\right)\right]\)
\(=\left(x^4+4x^2+4\right)-\left(x^4+4x^2-4x^2-16\right)\)
\(=x^4+4x^2+4-x^4-4x^2+4x^2+16\)
\(=4x^2+20\)
B. \(\left(x+2y\right)^2-\left(x-2y\right)^2\)
\(=\left(x^2+4xy+4y^2\right)-\left(x^2-4xy+4y^2\right)\)
\(=x^2+4xy+4y^2-x^2+4xy-4y^2\)
\(=6xy\)