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a) \(3y^2\left(2y-1\right)+y-y\left(1-y+y^2\right)-y^2+y \)
= \(6y^3-3y^2+y-y+y^2-y^3-y^2+y\)
= \(5y^3-3y^2+y\)
b)\(25x-4\left(3x-1\right)+\left(5-2x\right)7\)
= \(25x-12x+4+35-14x\)
= \(-x+39\)
c) \(11x-2\left(10x-1\right)-\left(4x-1\right)\left(-2\right)\)
= \(11x-\left(20x-2\right)-\left(-8x+2\right)\)
= \(11x-20x+2+8x-2\)
= \(-x\)
d) \(\left(\frac{1}{2x}\right)3-x\left(1-2x-\frac{1}{8x^2}\right)-x\left(x+\frac{1}{2}\right)\)
= \(\frac{3}{2x}-x+2x^2+\frac{x}{8x^2}-x^2-\frac{x}{2}\)
= \(\left(\frac{3}{2x}+\frac{1}{8x}-\frac{x}{2}\right)+x^2-x\)
= \(\left(\frac{12+1-4x^2}{8x}\right)+x^2-x\)
= \(\frac{13-4x^2}{8x}+\frac{8x^3}{8x}-\frac{8x^2}{8x}\)
= \(\frac{13-4x^2+8x^3-8x^2}{8x}\)
= \(\frac{8x^3-12x^2+13}{8x}\)
= x2 - \(\frac{3}{2}\)+\(\frac{13}{8x}\)
e) \(12\left(2-3x\right)+35x-\left(x+1\right)\left(-5\right)\)
= \(24-36x+35x-\left(-5x-5\right)\)
= \(24-36x+35x+5x+5\)
= 4x + 29
a. gọi phần đầu đấy là A nhá, để đỡ cần viết lại
A=...............
= (3x+5)2 + ( 3x-5)2 - 9x2 -4
= (9x2 +30x + 25 ) + ( 9x2 -30x+ 25 ) - 9x2 -4
= 9x2 +30x + 25 + 9x2 -30x+25-9x2 -4
= 9x2 + 46
sai thì thôi nhé. bạn nên kiểm tra lại
d. (2x-1)*(4x2 + 2x +1 ) - 8x*( x2 +1) - 5
= 8x3 -1 - 8x3 -8x-5
= -8x-6
= -2(4x+3)
sai nhé. bạn nên kiểm tra lại
a) \(x^2-y^2-5x-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
b) \(5x^3-5x^2y-10x^2+10xy\)
\(=\left(5x^3-5x^2y\right)-\left(10x^2-10xy\right)\)
\(=5x^2\left(x-y\right)-10x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x^2-10x\right)\)
\(=5x\left(x-y\right)\left(x-2\right)\)
c) \(x^3-2x^2-x+2\)
\(=\left(x^3-2x^2\right)-\left(x-2\right)\)
\(=x^2\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(-y^2+2xy-x^2+3x-3y\)
\(=-\left(y^2-2xy+x^2\right)+\left(3x-3y\right)\)
\(=-\left(y-x\right)^2+3\left(x-y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left[-\left(x-y\right)+3\right]\)
\(=\left(x-y\right)\left(-x+y+3\right)\)
g) \(4x^2-8x+3\)
\(=4x^2-6x-2x+3\)
\(=\left(4x^2-6x\right)-\left(2x-3\right)\)
\(=2x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(2x-3\right)\left(2x-1\right)\)
h) \(2x^2-5x-7\)
\(=2x^2+2x-7x-7\)
\(=\left(2x^2+2x\right)-\left(7x+7\right)\)
\(=2x\left(x+1\right)-7\left(x+1\right)\)
\(=\left(x+1\right)\left(2x-7\right)\)
k) \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
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
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