Ai giúp mình với !!!!!
x3.x2=28:23
Mình cần gấp
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 3:
a, (\(x\)+y+z)2
=((\(x\)+y) +z)2
= (\(x\) + y)2 + 2(\(x\) + y)z + z2
= \(x^2\) + 2\(xy\) + y2 + 2\(xz\) + 2yz + z2
=\(x^2\) + y2 + z2 + 2\(xy\) + 2\(xz\) + 2yz
b, (\(x-y\))(\(x^2\) + y2 + z2 - \(xy\) - yz - \(xz\))
= \(x^3\) + \(xy^2\) + \(xz^2\) - \(x^2\)y - \(xyz\) - \(x^2\)z - y3
Đến dây ta thấy xuất hiện \(x^3\) - y3 khác với đề bài, em xem lại đề bài nhé
Do x = -1 là nghiệm của phương trình
⇒ a - b - 1 - 2 = 0
⇒ a - b = 3
Tương tự ta có a + b = 1
Vậy a = 2 ; b = -1
a) (x - 140) : 7 = 33 - 23 . 3
(x - 140) : 7 = 27 - 8 . 3 = 27 - 24 = 3
x - 140 = 3 x 7 = 21
x = 21 + 140 = 161
b) x3 . x2 = 28 : 23
x5 = 25
=> x = 2
c) (x + 2) . ( x - 4) = 0
x = -2 hoặc 4
d) 3x-3 - 32 = 2 . 32 =
3x-3 - 9 = 2 . 9 = 18
3x-3 = 18 + 9 = 27
3x-3 = 33
=> x - 3 = 3
x = 3 + 3 = 6
\(\Leftrightarrow\left\{{}\begin{matrix}x^3+2y^2-4y+3=0\\2x^2+2x^2y^2-4y=0\left(1\right)\end{matrix}\right.\Rightarrow}x^3+2y^2-4y-2x^2-2x^2y^2+4y=0\Rightarrow x^3+1-2x^2y^2+2y^2-2x^2+2=0\Rightarrow\left(x+1\right)\left(x^2-x+1\right)-2y^2\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(x^2-x+1-2xy^2+2y^2-2x+2\right)=0\Rightarrow x=-1\)Thay x=-1 vào (1) ta được y2-2y+1=0⇒ (y-1)2=0⇒y-1=0⇒y=1
Do đó Q=x2+y2=(-1)2+12=2
[ (315 +372) x (3+7)] :338 + 1036
[687x 10] :1374
6870 : 1374
5
a) x2 ( x+ 2y) -x -2y
= x2 ( x+ 2y) -(x+2y)
= (x2-1)(x+2y)
= (x-1)(x+1)(x+2y)
b)3x2- 3y2 -2 (x-y)2
= 3(x2-y2) -2 (x-y)2
= 3(x-y)(x+y)-2(x-y)(x-y)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\\ =\left(x-y\right)\left(3x+3y-2x+2y\right)\\ =\left(x-y\right)\left(x+5y\right)\)
c) x2- 2x-4y2 - 4y
= (x2-4y2)-(2x+4y)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\\ =\left(x+2y\right)\left(x-2y-2\right)\)
d) x3 - 4x2 - 9x +36
= (x3+3x2)-(7x2+21x)+(12x+36)
= x2(x+3)-7x(x+3)+12(x+3)
=(x2-7x+12)(x+3)
\(=\left[\left(x^2-3x\right)-\left(4x-12\right)\right]\left(x+3\right)\\ =\left[x\left(x-3\right)-4\left(x-3\right)\right]\left(x+3\right)=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
sg8xcjvjbjbjnvnbnjbjbjbjbjjjgggggggggggggggggggggggggggggggggggggggggggggnnnnnnn mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
\(\left(2x+x^2\right)\left(x^2-3x+2\right)=0\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=2\end{matrix}\right.\\ A=\left\{-2;0;1;2\right\}\)
\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)
\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)
=>x5=25
=>x=2
ht
Vậy bạn có thể giải câu này đc ko??
3x+3x+2=270