Giúp tôi với
\(\left(8x-4x^2-1\right)\left(x^2+2x+1\right)=4\left(x^2+x-1\right)\)
d) \(^{ }4x\left(2x+3\right)-8x\left(x+4\right)\)
e) \(^{ }2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)
f) \(^{ }x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(4x\left(2x+3\right)-8x\left(x+4\right)\)
\(=8x^2+12x-8x^2-32x\)
=-20x
e: Ta có: \(2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)
\(=10x^2+4x+6x^2-2x-9x+3\)
\(=16x^2-7x+3\)
f: Ta có: \(x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3+4x^2+4x-x^3-3x^2-3x-1+3x^2-3\)
\(=4x^2+x-4\)
BT2: Tính giá trị biểu thức
\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)tại \(x=1\)
\(P=\left(2x-1\right)\left(4x^2-2x+1\right)-\left(1-2x\right)\left(1+2x+4x^2\right)\)tại \(x=10\)
\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)
\(M=\left(7-2x\right)\left[\left(2x\right)^2+2x\cdot7+7^2\right]-\left(64-8x^3\right)\)
\(M=\left[7^3-\left(2x\right)^3\right]-\left(64-8x^3\right)\)
\(M=343-8x^3-64+8x^3\)
\(M=279\)
Vậy M có giá trị 279 với mọi x
\(P=\left(2x-1\right)\left(4x^2-2x+1\right)-\left(1-2x\right)\left(1+2x+4x^2\right)\)
\(P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3\)
\(P=16x^3-8x^2+4x-2\)
Thay \(x=10\) vào P ta có:
\(P=16\cdot10^3-8\cdot10^2+4\cdot10-2=15238\)
Vậy P có giá trị 15238 tại x=10
a: M=343-8x^3-64+8x^3=279
b: P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3
=16x^3-8x^2+4x-2
=16*10^3-8*10^2+4*10-2=15238
Tìm x biết
1) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
2)\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x+1\right)-33\)
3)\(6x\left(3x+5\right)-2x\left(9x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)-17x^2=0\)
4)\(\left(x-1\right)\left(x+2\right)-\left(x-3\right)+5x-7=0\)
Giúp mình nha. Camon nhiều
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
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
Tìm x biết : \(\left(8x-4x^2+1\right)\left(x^2+2x+1\right)=4\left(x^2+x+1\right)\)
Giải Phương Trình \(\left(8x-4x^2-1\right)\left(x^2+2x+1\right)=4\left(x^2+x+1\right)\)
Bài 1. tính giá trị biểu thức.
a. \(5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x+2\right)\) với x = 15
b.\(5x\left(x-4y\right)-4y\left(y-5x\right)\) tại \(x=\dfrac{-1}{5}\) và \(y=\dfrac{-1}{2}\)
c.\(6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)với \(x=\dfrac{1}{2};y=2\)
giúp mik với mik đang cần gấp cảm ơn
Giải các bất phương trình, hệ phương trình
a) \(\dfrac{x^2-4x+3}{2x-3}\ge x-1\)
b) \(3x^2-\left|4x^2+x-5\right|>3\)
c)\(4x-\left|2x^2-8x-15\right|\le-1\)
d)\(x+3-\sqrt{21-4x-x^2}\ge0\)
e)\(\left\{{}\begin{matrix}x\left(x+5\right)< 4x+2\\\left(2x-1\right)\left(x+3\right)\ge4x\end{matrix}\right.\)
f)\(\dfrac{1}{x^2-5x+4}\le\dfrac{1}{x^2-7x+10}\)
Giải phương trình:
1, \(\left(x^2+x+1\right)\left(x^4+2x^3+7x^2+26x+37\right)=5\left(x+3\right)^3\)
2, \(\left(x+1\right)^3+\left(x+3\right)^3+6\left(x+1\right)\left(x+7\right)\left(x+3\right)=8\left(x+2\right)^3\)
3, \(x^3+\left(x-1\right)^3+3x\left(x-1\right)\left(x^4+x\right)=\left(2x-1\right)^3\)
4, \(\dfrac{\left(x+1\right)^3}{3x+1}+\dfrac{x^3+5x+2}{x^3+2x+1}=x+3\)
5, \(\dfrac{5x^3+x^2+x+1}{4x^2+1}+\dfrac{6\left(4x^2+1\right)}{x^3+x^2+1}=x+7\)
6, \(\left(x^2-4x+1\right)^3+\left(8x-x^2+4\right)^3+\left(x-5\right)^3=125x^3\)