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a) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x-84=30x-240-6x+84\)
\(\Rightarrow\left(72-84\right)-\left(20x+36x\right)=\left(30x-6x\right)-240+84\)
\(\Rightarrow-12-56=24x-56x\)
\(\Rightarrow-12+156=24x+56x\)
\(\Rightarrow144=80x\)
\(\Rightarrow x=144:80\)
\(\Rightarrow x=\frac{9}{5}\)
b) \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)
\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow15x+25-8x+12-5x-6x-36-1=0\)
\(\Rightarrow-4x=0\)
\(\Rightarrow-4.0\)
\(\Rightarrow x=0\)
\(ĐKXĐ:x\ne-1;x\ne\dfrac{1}{2}\)
Ta có : \(\dfrac{6x+5}{3x+3}=\dfrac{5-4x}{1-2x}\)
\(\Leftrightarrow\left(6x+5\right)\left(1-2x\right)=\left(5-4x\right)\left(3x+3\right)\)
\(\Leftrightarrow6x-12x^2+5-10x=15x+15-12x^2-12x\)
\(\Leftrightarrow5-4x-12x^2-15-3x+12x^2=0\)
\(\Leftrightarrow-10-7x=0\)
\(\Rightarrow x=\dfrac{-10}{7}\)
Tim x, bt:
a) 4.(18- 5x) - 12.( 3x-7) =15.(2x-16) - 6.(x+14)
b) 5.(3x+5) - 4.(2x-3) =5x + 3x(2x-12) +1
a) 4.(18- 5x) - 12.( 3x-7) =15.(2x-16) - 6.(x+14)
72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
-20x - 36x - 30x + 6x = -240 - 84 - 72 -84
-80x = -480
x= 6
4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6.(x+14)
4.18 - 4.5x - 12.3x + 12.7 = 15.2x - 15.16 - 6x - 6.14
72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
72 + 84 - 20x - 36x = 30x - 6x - 240 - 84
156 - 56x = 24x - 324
156 = 24x - 324 + 56x
156 = 80x - 324
80x - 324 = 156
80x = 156 + 324
80x = 480
x = 480:80
x = 6
câu b giải tương tự
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
Sửa đề bài 1 : k => x P/s : đề sai r :))
\(A=\left(3-2x\right)3x^2-8+\left(2x+5\right)\left(3x-2\right)-20x\)
\(=9x^2-6x^3-8+6x^2-4x+15x-10-20x=15x^2-6x^3-18-9x\)
Vậy biểu thức phụ thuộc biến x
\(B=\left(3-5x\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x+33-10x^2-55x-6x^2-14x-9x-21=-72x+12-16x^2\)
Vậy biểu thức phụ thuộc biến x
Bài 2 :
a, \(2x\left(x-1\right)-x^2+6=0\Leftrightarrow2x^2-2x-x^2+6=0\)
\(\Leftrightarrow x^2-2x+6=0\)( vô nghiệm )
b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^2-9-x\left(x^2-4\right)=15\Leftrightarrow x^2-9-x^3+12=15\)
\(\Leftrightarrow-x^3+x^2-12=0\Leftrightarrow x=2\)
a) \(3x\left(x-3\right)-5x\left(x+7\right)\)
\(=3x^2-9x-5x^2-35x\)
\(=-2x^2-44x\)
b) \(\dfrac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)-12\)
\(=2x^2-3x-2x^2+10x-12\)
\(=7x-12\)
a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow80x=480\Rightarrow x=6\)
b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow4x=0\Rightarrow x=0\)
c) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
a) \(3x+15=7\)
\(\Rightarrow3x=7-15\)
\(\Rightarrow3x=-8\)
\(\Rightarrow x=-8:3\)
\(\Rightarrow x=\frac{-8}{3}\)
Vậy x = \(\frac{-8}{3}\)
b) \(\left(x-3\right)\left(5-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\5-2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-3=0\\2x=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{5}{2}\end{cases}}\)
Vậy x = 3 hoặc x = \(\frac{5}{2}\)
_Chúc bạn học tốt_