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a,\(2x^2-8x=0\)
\(2x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b,\(B\left(x\right)=\left(2x^2-8x\right)-\left(3x+2x^2\right)\)
\(=2x^2-8x-3x-2x^2\)
=\(-11x\)
c,\(-11x=0\)
\(\Rightarrow x=0\)
\(A\left(x\right)=2x^2-8x\)
\(\Rightarrow2x^2-8x=0\)
\(\Rightarrow x\left(2x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=8\Rightarrow x=4\end{matrix}\right.\)
\(B\left(x\right)=-3x+2x^2\)
\(B\left(x\right)=2x^2-3x\)
\(2x^2-3x=0\)
\(\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
c) |2x + 3| - 4x < 9
Xét x ≥ -3/2
=> |2x + 3| - 4x < 9
<=> 2x + 3 - 4x < 9
<=> - 2x + 3 < 9
=> - 2x < 6
=> x < - 3
Xét x < -3/2 tương tự
b,xet 2 TH
TH1 3x-5=x+2
=>3x-x=2+5
=>2x=7
=>x=7/2
TH2 5-3x=x+2
=>-3x-x=5-2
=>-2x=3
=>x=-3/2
THẤY ĐÚNG THÌ
THANK
\(A\left(x\right)=8-5x+3x^2-15-3x+16=3x^2-8x+9\)
\(B\left(x\right)=5x-2x^2+4x-1-x^2-3x=-3x^2+6x-1\)
\(C\left(x\right)=B\left(x\right)-A\left(x\right)=\left(-3x^2+6x-1\right)-\left(3x^2-8x+9\right)\)
\(C\left(x\right)=-6x^2+14x-10\)
a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)
\(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)
Vậy x = 2
B, => 2x+5=3x-8
2x-3x=-8-5
-x=-13
=>x=13
hoặc 2x+5=-3x+8
2x+3x=8-5
5x=3
x=\(\frac{3}{5}\)
\(\left|x+1\right|+\left|x-5\right|=7.\)
\(Th1:x+1< 0;x-5< 0\)
\(x< -1;x< 5\Rightarrow x< -1\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(-\left(x+1\right)-\left(x-5\right)=7\)
\(-x-1-x+5=7\)
\(-2x+4=7\)
\(-2x=3\)
\(x=-1,5\left(tm\right)\)
\(Th2:x+1>0;x-5>0\)
\(x>-1;x>5\Rightarrow x>5\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(x+1+x-5=7\)
\(2x-4=7\)
\(2x=11\)
\(x=5,5\)\(\left(tm\right)\)
\(Th3:x+1\le0;x-5>0\)
\(x\le-1;x>5\)(không xảy ra)
\(Th4:x+1>0;x-5\le0\)
\(x>-1;x\le5\Rightarrow-1< x\le5\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(-\left(x+1\right)+x-5=7\)
\(-x-1+x-5=7\)( không xảy ra)
Vậy x = -1,5 hoặc x = 5,5
\(\)
a, |3x - 2| = x => 3x - 2 = x hoặc 3x - 2 = -x và x \(\ge\) 0 (vì |3x - 2| \(\ge\)0)
+) TH1: 3x - 2 = x => 3x - x = 2 => 2x = 2 => x = 1 (thỏa mãn)
+) TH2: 3x - 2 = -x => 3x + x = 2 => 4x = 2 => x = 0,5 (thỏa mãn)
Vậy x = 1 hoặc x = 0,5
b, |x - 2| = 2x + 1 => x - 2 = 2x + 1 và 2x + 1 \(\ge\)0 (vì |x - 2| \(\ge\)0) => x \(\ge\) -0,5 hoặc x - 2 = -(2x + 1) và -(2x + 1) \(\ge\)0 (vì |x - 2| \(\ge\)0) => x \(\ge\) -0,5
+)TH1: x - 2 = 2x + 1 => x - 2x = 1 + 2 => -x = 3 => x = -3 (loại)
+)TH2: x - 2 = -(2x + 1) => x - 2 = -2x - 1 => x + 2x = -1 + 2 => 3x = 1 => x = \(\frac{1}{3}\)(chọn)
Vậy: x = \(\frac{1}{3}\)