Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: ta có: \(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)
\(\Leftrightarrow2x^2+4x-5x-10-2x^2+2x=15\)
\(\Leftrightarrow x=25\)
b: Ta có: \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)
c: Ta có: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)
\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)
\(\Leftrightarrow-9x=1\)
hay \(x=-\dfrac{1}{9}\)
1: \(\left(x-3\right)\left(2x-5\right)-3x\left(x+4\right)\)
\(=2x^2-5x-6x+15-3x^2-12x\)
\(=-x^2-23x+15\)
2: \(\left(\dfrac{1}{2}x+5\right)\left(2x-\dfrac{1}{5}\right)\)
\(=x^2-\dfrac{1}{10}x+10x-1\)
\(=x^2+\dfrac{99}{10}x-1\)
a: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)
A) 3x² - x(3x - 5) = 9
3x² - 3x² + 5x = 9
5x = 9
x = 9/5
--------------------
B) 5x² + 9x - 2 = 0
5x² + 10x - x - 2 = 0
(5x² + 10x) - (x + 2) = 0
5x(x + 2) - (x + 2) = 0
(x + 2)(5x - 1) = 0
x + 2 = 0 hoặc 5x - 1 = 0
*) x + 2 = 0
x = -2
*) 5x - 1 = 0
5x = 1
x = 1/5
Vậy x = -2; x = 1/5
---------------------
D) 4(5 - 3x) = 5x - 5
20 - 12x = 5x - 5
-12x - 5x = -5 - 20
-17x = -25
x = 25/17
--------------------
E) 2x² - 11x + 14 = 0
2x² - 4x - 7x + 14 = 0
(2x² - 4x) - (7x - 14) = 0
2x(x - 2) - 7(x - 2) = 0
(x - 2)(2x - 7) = 0
x - 2 = 0 hoặc 2x - 7 = 0
*) x - 2 = 0
x = 2
*) 2x - 7 = 0
2x = 7
x = 7/2
Vậy x = 2; x = 7/2
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
a)\(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)
\(=\left(x^2-9\right)\left(x+2\right)-\left(x^3-3x-x^2+3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)
\(=x^3+2x^2-9x-18-x^3+x^2+3x-3-5x^3-40x^2-80x-x^2+10x-25\)
\(=-5x^3-38x^2-76x-46\)
b)\(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)
\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2+10x+25\right)-\left(x^2-2x+1\right)\)
\(=2x^3-16x^2+32x-\left(x^3+5x^2-4x-20\right)+2x^2+20x+50-x^2+2x-1\)
\(=x^3-20x^2+58x+69\)
c)\(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3-48x^2-36x-\left(2x^3-x^2-18x+9\right)\)
\(=-18x^3-46x^2-8x+16\).
a) Ta có: \(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)
\(=\left(x^2-9\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)
\(=x^3+2x^2-9x-18-\left(x^3-3x-x^2+3\right)-5x^3-40x^2-80x-x^2+10x-25\)
\(=-4x^3-39x^2-79x-43-x^3+3x+x^2-3\)
\(=-5x^3-38x^2-76x-46\)
b) Ta có: \(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)
\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2x^2+20x+50-x^2+2x-1\)
\(=2x^3-16x^2+32x-x^3+4x-5x^2+20+x^2+22x+49\)
\(=x^3-20x^2+56x+49\)
c) Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x-3\right)\left(x+3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3+48x-36x-2x^3+18x+x^2-9\)
\(=-18x^3+2x^2+40x+16\)
\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)
\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)
Vậy \(x=\dfrac{-255}{2}\)
\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)
(x-1)(5-2x)+2(x+5)(2x-5)
=-(x-1)(2x-5)+2(x+5)(2x-5)
=(2x-5)(x+1+2x+10)
=(2x-5)(3x+11)
TFboys_Lê Phương Thảo Bạn quên thêm dấu trừ vào vế tích thứ nhất kìa