-3 (2x + 5 ) - 16 < -4 (3 - 2x)
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- \(15+4x< 2x-145\)
\(\Rightarrow15+4x-2x+145< 0\)
\(\Rightarrow\left(15+145\right)+\left(4x-2x\right)< 0\)
\(\Rightarrow160+2x< 0\)
\(\Rightarrow2x< 0-160\)
\(\Rightarrow2x< -160\)
\(\Rightarrow x< \left(-160\right)\div2\)
\(\Rightarrow x< -80\)
Vậy x < -80
- \(-3\left(2x+5\right)-16< -4\left(3-2x\right)\)
\(\Rightarrow\left(-3\right)\times2x+\left(-3\right)\times5-16< \left(-4\right)\times3-\left(-4\right)\times2x\)
\(\Rightarrow-6x+\left(-15\right)-16< \left(-12\right)-\left(-8x\right)\)
\(\Rightarrow-6x-15-16< \left(-12\right)+8x\)
\(\Rightarrow-6x-31< \left(-12\right)+8x\)
\(\Rightarrow-6x-31+12-8x< 0\)
\(\Rightarrow-6x-8x-31+12< 0\)
\(\Rightarrow\left(-6x-8x\right)-19< 0\)
\(\Rightarrow-14x-19< 0\)
\(\Rightarrow-14x< 0+19\)
\(\Rightarrow-14x< 19\)
\(\Rightarrow x< 19\div\left(-14\right)\)
\(\Rightarrow x< -\frac{19}{14}\)
Vậy \(x< -\frac{19}{14}\)
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) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
Bạn nên viết đề bằng công thức toán và ghi đầy đủ yêu cầu đề để mọi người hiểu đề của bạn hơn nhé.
a: =>3|x-4|=16-2x
TH1: x>=4
=>3x-12=16-2x
=>5x=28
=>x=28/5(nhận)
TH2: x<4
=>12-3x=16-2x
=>-x=4
=>x=-4(nhận)
b: =>|2x-5|=7+4x+4=4x+11
TH1: x>=5/2
=>4x+11=2x-5
=>2x=-16
=>x=-8(loại)
TH2: x<5/2
=>4x+1=5-2x
=>6x=4
=>x=2/3(nhận)
-3 (2x + 5 ) - 16 < -4 (3 - 2x) => -6x-15-16 < -12 + 8x
=> -6x - 31 < -12 + 8x
=> -6x - 31 +12 - 8x <0
=> -14x - 19 < 0
=> -14x < 19
=> x>-19/14
Chúc bạn học tốt!