![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow12x^2-42x-\left(8x-28\right)+\left(3x-12x^2\right)-\left(5-20x\right)=-31\)
\(\Leftrightarrow12x^2-42x-8x+28+3x-12x^2-5+20x=-31\)
\(\Leftrightarrow\left(12x^2-12x^2\right)-x\left(42+8\right)+x\left(3+20\right)+28-5=-31\)
\(\Leftrightarrow0-50x+23x+23=-31\)
\(\Leftrightarrow-27x=-31-23\)
\(\Leftrightarrow-27x=-54\)
\(\Leftrightarrow x=\left(-54\right)\div\left(-27\right)\)
\(\Leftrightarrow x=2\)
Vậy x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(2x-3\right)\left(6x-4\right)-\left(3x+1\right)\left(5x-5\right)=31\)
\(\Leftrightarrow12x^2-8x-18x+12-\left(15x^2-15x+5x-5\right)=31\)
\(\Leftrightarrow12x^2-20x+12-15x^2+10x+5-31=0\)
\(\Leftrightarrow-3x^2-10x-14=0\)
\(\Leftrightarrow3x^2+10x+14=0\)
\(\text{Δ}=10^2-4\cdot3\cdot14=100-168=-68< 0\)
Do đó: Phương trình vô nghiệm
b: \(\left(2x-3\right)\left(4x-5\right)-\left(8x+1\right)\left(x-2\right)=12\)
\(\Leftrightarrow8x^2-10x-12x+15-\left(8x^2-16x+x-2\right)=12\)
\(\Leftrightarrow8x^2-22x+15-8x^2+15x+2=12\)
=>-7x+17=12
=>-7x=-5
hay x=5/7
![](https://rs.olm.vn/images/avt/0.png?1311)
a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)
=\(-x^5+2x^4-4x^2-1\)
f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
=\(3x^5-10x^4-13\)
b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)
=\(x^4+9x^3-11x^2+7x-2\)
f(x)-g(x)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)
=\(9x^4+5x^3-x^2-x-12\)
a )
\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
\(\left(6x-4\right)\left(2x-7\right)+\left(3x+5\right)\left(x-4x\right)=-31\\ \Leftrightarrow12x^2-50x+28+3x^2-12x^2+5x-20x+31=0\)
\(\Leftrightarrow3x^2-65x+59=0\)
\(\Leftrightarrow x^2-\dfrac{65}{3}x+\dfrac{59}{3}=0\\ \Leftrightarrow x^2-2.\dfrac{65}{6}x+\left(\dfrac{65}{6}\right)^2=-\dfrac{59}{3}+\left(\dfrac{65}{6}\right)^2\)
\(\Leftrightarrow\left(x-\dfrac{65}{6}\right)^2=\dfrac{3517}{36}\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{65}{6}=\dfrac{\sqrt{3517}}{6}\\\dfrac{65}{6}-x=\dfrac{\sqrt{3517}}{6}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt{3517}+65}{6}\\x=\dfrac{-\sqrt{3517}+65}{6}\end{matrix}\right.\)
vậy...