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1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ
2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
1. (2x - 3) . (2x+3) - 4 . (x+ 2)2 = 6
[ ( 2x )2 - 32 ] - 4 . ( x2 + 2.x.2 + 22) = 6
4x2 - 9 - 4 . ( x2 + 4x + 4) = 6
4x2 - 9 - 4x2 - 16x - 16 = 6
-16x -25 = 6
x = \(-\dfrac{31}{16}\)
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
a, \(\left(3x+2\right)^2-\left(2x-1\right)\left(2x+1\right)=5\left(x-2\right)^2\)
\(\Rightarrow9x^2+12x+4-\left(4x^2-1\right)=5\left(x^2-4x+4\right)\)
\(\Rightarrow9x^2+12x+4-4x^2-1=5x^2-20x+20\)
\(\Rightarrow9x^2-4x^2-5x^2+12x+20x=20+1-4\)
\(\Rightarrow32x=17\Rightarrow x=\dfrac{17}{32}\)
b, \(\left(x+2\right)^2-\left(x+3\right)\left(x-1\right)=5x\)
\(\Rightarrow x^2+4x+4-\left(x^2-x+3x-3\right)=5x\)
\(\Rightarrow x^2+4x+4-x^2+x-3x+3-5x=0\)
\(\Rightarrow-3x=-3-4\Rightarrow-3x=-7\Rightarrow x=\dfrac{7}{3}\)
c, \(\left(3x-1\right)\left(x-3\right)+\left(x-2\right)^2=\left(2x-5\right)^2\)
\(\Rightarrow3x^2-9x-x+3+x^2-4x+4=4x^2-20x+25\)
\(\Rightarrow3x^2+x^2-4x^2-9x-x-4x+20x=25-3-4\)
\(\Rightarrow6x=18\Rightarrow x=3\)
Chúc bạn học tốt!!!
a)\((x^2- 4).(x^2 - 10) = 72 Đặt x^2 - 7 = a(1), ta có (a+3)(a-3)=72 a^2-9=72 a^2=81 a=+-9 xét 2 trường hợp a = 9 và -9 khi thay vào (1) ta có..... tự lm nốt nha \)
b) nhóm x+1 vs x+4 và x+2 vs x+3 ta sẽ có (x2+5x+4)(x2+5x+6)(x+5)=40
a) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2=2x-3\\ \Leftrightarrow x^2-25-x^2-6x-9-2x+3=0\\ \Leftrightarrow-31-8x=0\\ \Leftrightarrow8x=-31\\ \Leftrightarrow x=\dfrac{-31}{8}\)
b)\(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2\\ \Leftrightarrow4x^2+12x+9+x^2-1-5\left(x^2+4x+4\right)=0\\ \Leftrightarrow5x^2+12x+8-5x^2-20x-20=0\\ \Leftrightarrow-8x-12=0\\ \Leftrightarrow-8x=12\\ \Leftrightarrow x=\dfrac{-3}{2}\)
1: \(=x^3-6x^2+12x-8-8x^3-36x^2-54x-27+7\left(x-1\right)^3\)
\(=-7x^3-42x^2-42x-35+7x^3-21x^2+21x-7\)
\(=-63x^2-21x-42\)
2: \(=x^3+125-\left(x^3-8\right)=125+8=133\)
3: \(=8x^3-27-8x^3-12x^2-6x-1=-12x^2-6x-28\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{4}{3}-\dfrac{5}{4}x+\dfrac{5}{4}=\dfrac{15}{2}-\dfrac{3}{2}x-\dfrac{3}{2}\left(2x+3\right)\)
\(\Leftrightarrow x\cdot\dfrac{-7}{12}+\dfrac{31}{12}=\dfrac{-15}{2}x+3\)
=>83/12x=5/12
hay x=5/83
1.\(\left(x-5\right).\left(x+5\right)-\left(x+3\right)^2=2x-3\)
\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9-2x+3=0\)
\(\Leftrightarrow-8x-31=0\)
\(\Leftrightarrow x=\dfrac{-31}{8}\)
\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-\left(x^3-5^3\right)=\left(x^3+2^3\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-x^3+5^3=x^3+2^3-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x^3-12x^2+48x-64\right)-x^3+5^3=x^3+2^3-\left(x^3+12x^2+48x+64\right)\)
\(\Leftrightarrow x^3-12x^2+48x-64-x^3+5^3=x^3+2^3-x^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-64+5^3=2^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-61=-12x^2-48x-56\)
\(\Leftrightarrow96x=-117\)
\(\Leftrightarrow x=\dfrac{-117}{96}=\dfrac{-39}{32}\)