\(\left(2x-1\right)^2+\left(3x+2\right)^2-2\left(2x-1\right)\left(3x+2\right)\)
Mong mọi ng giải hộ mik bài này , mik đg cần gấp. Thanks ạ
\(\left(2x-1\right)^2+\left(3x+2\right)^2-2\left(2x-1\right)\left(3x+2\right)\)
Mong mọi ng giải hộ mik bài này , mik đg cần gấp. Thanks ạ
Bạn có thể thấy 2x-1 là a , 3x+2 là b thì 2.(2x-1)(3x+2)=2ab
nên phương trình trên có thể dùng bình phương 1 tổng
\(\left(2x-1\right)^2+\left(3x+2\right)^2-2.\left(2x-1\right).\left(3x+2\right)=\left[\left(2x-1\right)-\left(3x+2\right)\right]^2\)
\(=\left(2x-1-3x-2\right)^2=\left(-x-3\right)^2=\left(x+3^2\right)\)
cái cuối là (x+3)2 nhé, không phải (x+32) đâu
Giải các phương trình sau:
1, \(\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\)
2, \(\left(x-2\right)\left(2x-1\right)=x^2-2x\)
3, \(3x^2-4x+1=0\)
4, \(\left|2x-4\right|=0\)
5, \(\left|3x+2\right|=4\)
6, \(\left|2x-5\right|=\left|-x+2\right|\)
*Giúp mình với mình đg cần gấp ạ T_T
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
Mọi người dành thời gian giải hộ mình bài toán với:
B2: Giải các PT sau:
d) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x+1\right)\left(7x-10\right)\)
e) \(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
f) \(x\left(2x-7\right)-4x+14=0\)
g) \(3x-15=2x\left(x-5\right)\)
h) \(\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
i) \(0,5x\left(x-3\right)=\left(x-3\right)\left(1,5x-1\right)\)
j) \(\left(2x^2+1\right)\left(4x-3\right)=\left(x-12\right)\left(2x^2+1\right)\)
k) \(x\left(2x-9\right)=3x\left(x-5\right)\)
Các Pro giải giúp mik với :(
e sẽ cố gắng !!!
\(3x-15=2x\left(x-5\right)\)
\(3x-15=2x^2-10x\)
\(3x-15-2x^2+10x=0\)
\(13x-15-2x^2=0\)
\(x\left(13-2x\right)-15=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\13-2x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\-2-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\2x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
\(f,x\left(2x-7\right)-4x+14=0\)
\(2x^2-7x-4x+14=0\)
\(2x^2-11x+14=0\)
\(x\left(2x-11\right)=-14\)
\(\Rightarrow\orbr{\begin{cases}x=-14\\2x-11=-14\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-14\\x=-\frac{3}{2}\end{cases}}}\)
\(e,\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(3x+36-8x=x^2+4x+4\)
\(-5x+36-x^2-4x-4=0\)
\(-9x+32-x^2=0\)
\(x\left(-9-x\right)+32=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\23-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=23\end{cases}}}\)
chúc cj hay a hc tốt
tìm x biết
a) \(\left(x-2\right)^3\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
b)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
c)\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
d)\(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
Giúp mk vs đc k ạ mk đg cần gấp
\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8=7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
Bài 1: Rút gọn:
\(A=\left(2a^2+2a+1\right).\left(2a^2-2a+1\right)-\left(2a^2+1\right)^2\)
\(B=5.\left(2x-1\right)^2+4.\left(x-1\right).\left(x+3\right)-2.\left(5-3x\right)^2\)
GIÚP MIK VS!!! MIK ĐAG CẦN GẤP
giải hệ phương trình x^3+y^2+xy(x+y)=13 và x^2*y^2*(x^2+y^2)=468
\(\left\{\begin{matrix}\sqrt{3x-1}+4\left(2x+1\right)=\sqrt{y-1}+3y\\\left(x+y\right)\left(2x-y\right)+4=-6-3y\end{matrix}\right.\)
Mọi người giải giúp mình với ạ! đang cần gấp...Thanks!
\(\left(2x-x^2\right)3\left(x^2+3x+1\right)\)
giải hộ em vs ạ
\(=3x^2\left(x^2+3x+1\right)\)
=3x^4+9x^3+3x^2
tìm GTNN biết \(\left(x^z+1\right)^2+\left(y^z+2\right)^4-2\)
hộ mik vs mik đg cần gấp
x^2+1>=1
=>(x^2+1)^2>=1
y^2+2>=2
=>(y^2+2)^4>=16
=>(x^2+1)^2+(y^2+2)^4>=17
=>(x^2+1)^2+(y^2+2)^4-2>=15
Dấu = xảy ra khi x=y=0
Mọi người ơi, giúp mình với mình đang cần gấp
thank mọi người
1 thực hiện phép nhân
1)\(-3\left(-x+3\right)\)
2)\(-5x^3\left(-3x+5\right)\)
3)\(-2x\left(-2x-6\right)\)
4)\(-3x^3\left(-2x-12\right)\)
5)\(-10x\left(-5x+2\right)\)
6)\(-2x^2\left(-3x^3+4x^2-5\right)\)
7)\(-x^2\left(-2x^3-x+3\right)\)
8)\(-2^3\left(-2-6\right)\)
9)\(\left(-x-3\right)\left(x+2\right)\)
10)\(\left(2x-3\right)\left(5-x\right)\)
11)\(\left(-x+6\right)\left(-x-2\right)\)
12)\(\left(3x-1\right)\left(-3-2x\right)\)
13)\(\left(-5x-3\right)\left(2-x\right)\)
14) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)
15) \(\left(x-3\right)\left(1-2x-5y\right)\)
16)\(\left(x-2\right)\left(x^2+4\right)\left(x+2\right)\)
17)\(\left(-3+1\right)\left(9x^2+1\right)\left(-3x-1\right)\)
18)\(-\left(2x+1\right)\left(2-1\right)\left(x+1\right)\)
19)\(\left(2x^2-3x+5\right)\left(x^2-8x+2\right)\)
20)\(\left(3x^2y-6xy+9x\right)\left(-xy\right)\)
1. -3(-x+3)
= 3x - 6
2. -5x3 (-3x + 5)
= 15x4 - 25x3
3. -2x (-2x - 6)
= 4x2 + 12x