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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
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Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
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\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{4}{3}-\dfrac{5}{4}x+\dfrac{5}{4}=\dfrac{15}{2}-\dfrac{3}{2}x-\dfrac{3}{2}\left(2x+3\right)\)
\(\Leftrightarrow x\cdot\dfrac{-7}{12}+\dfrac{31}{12}=\dfrac{-15}{2}x+3\)
=>83/12x=5/12
hay x=5/83
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Câu 1:
\(\Leftrightarrow6x-18-8x-4-2x+8=4-3\left(2x+1\right)+5\left(2x-1\right)\)
=>-4x-14=4-6x-3+10x-5
=>-4x-14=4x-4
=>-8x=10
hay x=-5/4
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\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
\(=\dfrac{3^{29}.2^3}{2^2.3^{28}}\)
\(=\dfrac{3.2}{1.1}=6\)
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+) Nếu x-2=0 =>x=2
+) Nếu x2-4=0 => x=2 hoặc x= -2
+) Nếu 3x-9=0 => x=3
+) Nếu x3+8=0 =>x= - 2
Vậy để biểu thức bằng 0 thì x=2 ; x=-2 ; x=3
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1: \(\Leftrightarrow x^2-4x+4-2\left(x^2+2x+1\right)=\left(2x+1\right)\left(1-3x\right)+2x\left(x-1\right)\)
\(\Leftrightarrow x^2-4x+4-2x^2-4x-2=\left(2x-6x^2+1-3x\right)+2x^2-2x\)
\(\Leftrightarrow-x^2-8x+2=-6x^2-x+1+2x^2-2x\)
\(\Leftrightarrow-x^2-8x+2=-4x^2-3x+1\)
\(\Leftrightarrow3x^2-5x+1=0\)
\(\Delta=\left(-5\right)^2-4\cdot3\cdot1=25-12=13>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{5-\sqrt{13}}{6}\\x_2=\dfrac{5+\sqrt{13}}{6}\end{matrix}\right.\)
2x(X−3)=3⋅(3xX−9)
=> 2xX - 2x3 = 3 . 3xX - 3. 9
=> 2x2 - 6 = 9x2 - 27
=> 2x2 - 2.3 = 9x2 - 9.3
=> 2 . (x2 - 3) = 9 . ( x2 - 3)
=> 0 = 7 . ( x2 - 3)
=> 0 = x2 - 3
=> 0 + 3 = x2
=> 3 = x2