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\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
a) Ta có: \(\left|2x-1\right|=\left|2x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2x+3\left(loại\right)\\2x-1=-2x-3\end{matrix}\right.\Leftrightarrow2x+2x=-3+1\)
\(\Leftrightarrow4x=-2\)
hay \(x=-\dfrac{1}{2}\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
Làm tiếp nè :
2) / 2x + 4/ = 2x - 5
Do : / 2x + 4 / ≥ 0 ∀x
⇒ 2x - 5 ≥ 0
⇔ x ≥ \(\dfrac{5}{2}\)
Bình phương hai vế của phương trình , ta có :
( 2x + 4)2 = ( 2x - 5)2
⇔ ( 2x + 4)2 - ( 2x - 5)2 = 0
⇔ ( 2x + 4 - 2x + 5)( 2x + 4 + 2x - 5) = 0
⇔ 9( 4x - 1) = 0
⇔ x = \(\dfrac{1}{4}\) ( KTM)
Vậy , phương trình vô nghiệm .
3) / x + 3/ = 3x - 1
Do : / x + 3 / ≥ 0 ∀x
⇒ 3x - 1 ≥ 0
⇔ x ≥ \(\dfrac{1}{3}\)
Bình phương hai vế của phương trình , ta có :
( x + 3)2 = ( 3x - 1)2
⇔ ( x + 3)2 - ( 3x - 1)2 = 0
⇔ ( x + 3 - 3x + 1)( x + 3 + 3x - 1) = 0
⇔ ( 4 - 2x)( 4x + 2) = 0
⇔ x = 2 (TM) hoặc x = \(\dfrac{-1}{2}\) ( KTM)
KL......
4) / x - 4/ + 3x = 5
⇔ / x - 4/ = 5 - 3x
Do : / x - 4/ ≥ 0 ∀x
⇒ 5 - 3x ≥ 0
⇔ x ≤ \(\dfrac{-5}{3}\)
Bình phương cả hai vế của phương trình , ta có :
( x - 4)2 = ( 5 - 3x)2
⇔ ( x - 4)2 - ( 5 - 3x)2 = 0
⇔ ( x - 4 - 5 + 3x)( x - 4 + 5 - 3x) = 0
⇔ ( 4x - 9)( 1 - 2x) = 0
⇔ x = \(\dfrac{9}{4}\) ( KTM) hoặc x = \(\dfrac{1}{2}\) ( KTM)
KL......
Làm tương tự với các phần khác nha
1)\(\left|4x\right|=3x+12\)
\(\Leftrightarrow4.\left|x\right|=3x+12\\ \Leftrightarrow4.\left|x\right|-3x=12\)
\(TH1:4x-3x=12\left(x\ge0\right)\\\Leftrightarrow x=12\left(TM\right) \)
\(TH2:4.\left(-x\right)-3x=12\left(x< 0\right)\\ \Leftrightarrow-7x=12\\ \Leftrightarrow x=-\dfrac{12}{7}\left(TM\right)\)
Vậy tập nghiệm của PT: \(S=\left\{12;-\dfrac{12}{7}\right\}\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)
a) \(\Leftrightarrow2\left|3x-1\right|=\dfrac{4}{5}\)
\(\Leftrightarrow\left|3x-1\right|=\dfrac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\\3x-1=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{1}{5}\end{matrix}\right.\)
b)TH1: \(x\ge3\)
\(\Leftrightarrow x+5+x-3=9\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\)
TH2: \(-5\le x< 3\)
\(\Leftrightarrow x+5-x+3=9\Leftrightarrow8=9\left(VLý\right)\)
TH3: \(x< -5\)
\(\Leftrightarrow-x-5-x+3=9\Leftrightarrow2x=-11\Leftrightarrow x=-\dfrac{11}{2}\left(tm\right)\)
\(a,2.|3x-1|-\dfrac{3}{4}=\dfrac{1}{20}\)
\(2.|3x-1|=\dfrac{1}{20}+\dfrac{3}{4}\)
\(2.|3x-1|=\dfrac{4}{5}\)
\(|3x-1|=\dfrac{4}{5}:2\)
\(|3x-1|=\dfrac{2}{5}\)
\(\Rightarrow3x-1=\pm\dfrac{2}{5}\)
\(3x-1=\dfrac{2}{5}\)
\(3x=\dfrac{2}{5}+1\)
\(3x=\dfrac{7}{5}\)
\(x=\dfrac{7}{5}:3\)
\(x=\dfrac{7}{15}\)
\(3x-1=-\dfrac{2}{5}\)
\(3x=-\dfrac{2}{5}+1\)
\(3x=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:3\)
\(x=\dfrac{1}{5}\)
a, \(\left|x-1\right|=3x+2\)ĐK : x >= -2/3
TH1 : \(x-1=3x+2\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)( ktm )
TH2 : \(x-1=-3x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)( tm )
b, \(\left|5x\right|=x-12\)ĐK : x > = 12
TH1 : \(5x=x-12\Leftrightarrow4x=-12\Leftrightarrow x=-3\)( ktm )
TH2 : \(5x=12-x\Leftrightarrow6x=12\Leftrightarrow x=2\)( ktm )
c, \(\left|7-x\right|=5x+1\)ĐK : x >= -1/5
TH1 : \(7-x=5x+1\Leftrightarrow6x=-6\Leftrightarrow x=-1\)( ktm )
TH2 : \(7-x=5x+1\Leftrightarrow6x=6\Leftrightarrow x=1\)( tm )
a) Điều kiện : \(x\ge-\frac{2}{3}\)
Xét :
\(\Rightarrow x-1< 0\)
\(\Rightarrow\left|x-1\right|=1-x\)
\(\Rightarrow1-x=3x+2\)
\(\Rightarrow4x=-1\)
\(\Rightarrow x=-\frac{1}{4}\left(TM\right)\)
\(\Rightarrow x-1\ge0\)
\(\Rightarrow\left|x-1\right|=x-1\)
\(\Rightarrow x-1=3x+2\)
\(\Rightarrow2x=-3\)
\(\Rightarrow x=-\frac{3}{2}\left(KTM\right)\)
Vậy \(x=-\frac{1}{4}\)
b) Điều kiện : \(x\ge12\)
\(\Rightarrow5x>0\)
\(\Rightarrow\left|5x\right|=5x\)
\(\Rightarrow5x=x-12\)
\(\Rightarrow4x=-12\)
\(\Rightarrow x=-3\left(KTM\right)\)
Vậy không có x
c)Điều kiện : \(x\ge-\frac{1}{5}\)
Xét :
\(\Rightarrow7-x\ge0\)
\(\Rightarrow\left|7-x\right|=7-x\)
\(\Rightarrow7-x=5x+1\)
\(\Rightarrow6x=6\)
\(\Rightarrow x=1\left(TM\right)\)
\(\Rightarrow7-x< 0\)
\(\Rightarrow\left|7-x\right|=x-7\)
\(\Rightarrow x-7=5x+1\)
\(\Rightarrow4x=-8\)
\(\Rightarrow x=-2\left(KTM\right)\)
Vậy \(x=1\)