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1: Ta có: |2x-3|=|x+5|
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x+5\\2x-3=-x-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-3-x-5=0\\2x-3+x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{8;\frac{-2}{3}\right\}\)
2: Ta có: |4-2x|=|3x|
\(\Leftrightarrow\left[{}\begin{matrix}4-2x=3x\\4-2x=-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4-2x-3x=0\\4-2x+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x+4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x=-4\\x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=-4\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{4}{5};-4\right\}\)
3: Ta có: |4x-5|-|2x+1|=0
\(\Leftrightarrow\left|4x-5\right|=\left|2x+1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-5=2x+1\\4x-5=-2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-5-2x-1=0\\4x-5+2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\6x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\6x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{3;\frac{2}{3}\right\}\)
4: Ta có: \(\left|0.5x-2\right|-\left|x+\frac{2}{3}\right|=0\)
\(\Leftrightarrow\left|0.5x-2\right|=\left|x+\frac{2}{3}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x-2=x+\frac{2}{3}\\\frac{1}{2}x-2=-x-\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x-2-x-\frac{2}{3}=0\\\frac{1}{2}x-2+x+\frac{2}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{-1}{2}x-\frac{8}{3}=0\\\frac{3}{2}x-\frac{4}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{-1}{2}x=\frac{8}{3}\\\frac{3}{2}x=\frac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}:\frac{-1}{2}=\frac{8}{3}\cdot\left(-2\right)=\frac{-16}{3}\\x=\frac{4}{3}:\frac{3}{2}=\frac{4}{3}\cdot\frac{2}{3}=\frac{8}{9}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{-16}{3};\frac{8}{9}\right\}\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
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: \(S=\left\{12;-\dfrac{12}{7}\right\}\)
a) Ta có: \(\left|2x-5\right|\ge0\forall x\)
\(\left|3y+1\right|\ge0\forall y\)
Do đó: \(\left|2x-5\right|+\left|3y+1\right|\ge0\forall x,y\)
mà \(\left|2x-5\right|+\left|3y+1\right|=0\)
nên \(\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|3y+1\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x=\frac{5}{2}\) và \(y=\frac{-1}{3}\)
b) Ta có: \(\left|3x-4\right|\ge0\forall x\)
\(\left|3y-5\right|\ge0\forall y\)
Do đó: \(\left|3x-4\right|+\left|3y-5\right|\ge0\forall x,y\)
mà \(\left|3x-4\right|+\left|3y-5\right|=0\)
nên \(\left\{{}\begin{matrix}\left|3x-4\right|=0\\\left|3y-5\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-4=0\\3y-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=\frac{5}{3}\end{matrix}\right.\)
Vậy: \(x=\frac{4}{3}\) và \(y=\frac{5}{3}\)
c) Ta có: |16-|x||≥0∀x
\(\left|5y-2\right|\ge0\forall y\)
Do đó: |16-|x||+|5y-2|≥0∀x,y
mà |16-|x||+|5y-2|=0
nên \(\left\{{}\begin{matrix}\text{|16-|x||}=0\\\left|5y-2\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}16-\left|x\right|=0\\5y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|=16\\5y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{16;-16\right\}\\y=\frac{2}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{16;-16\right\}\) và \(y=\frac{2}{5}\)
có |2x-5| luôn \(\ge0\forall x\in Q\)
cũng có \(\left|3y+1\right|\ge0\forall y\in Q\)
=> \(\left|2x-5\right|+\left|3y-1\right|\ge0\forall x;y\in Q\)
=>\(\hept{\begin{cases}2x-5=0\\3y-1=0\end{cases}}\)<=> \(\hept{\begin{cases}2x=5\\3y=1\end{cases}}\)<=> \(\hept{\begin{cases}x=\frac{2}{5}\\y=\frac{1}{3}\end{cases}}\)
vậy \(x=\frac{2}{5};y=\frac{1}{3}\)
em nhớ là phải dùng ngoặc nhọn như trên nhé! Nếu không sẽ sai đấy!
3 câu còn lại cũng tương tự
\(\frac{2^{4-x}}{16^5}=32^6\)
=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)
=> \(2^{4-x}=2^{30}.2^{20}\)
=> \(2^{4-x}=2^{50}\)
=> 4 - x = 50
=> x = 4 - 50 = -46
\(\frac{3^{2x+3}}{9^3}=9^{14}\)
=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)
=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)
=> \(3^{2x+3}=3^{28}.3^6\)
=> \(3^{2x+3}=3^{34}\)
=> 2x + 3 = 34
=> 2x = 34 - 3
=> 2x = 31
=> x = 31/2
\(\left(3-4x\right)^2=25=5^2\)
\(\Rightarrow3-4x=5\)
\(\Rightarrow4x=3-5=-2\Rightarrow x=-\frac{1}{2}\)
\(\left(2x-\frac{1}{4}\right)^2=16=4^2\)
\(\Rightarrow2x-\frac{1}{4}=4\Rightarrow2x=4+\frac{1}{4}=\frac{17}{4}\)
\(\Rightarrow x=\frac{17}{4}:2=\frac{17}{4}.\frac{1}{2}=\frac{17}{8}\)
Đề số 3 bị sai.
\(\left(2x+5\right)^2=0\Rightarrow2x+5=0\Rightarrow2x=-5\Rightarrow x=-\frac{5}{2}\)
(3-4x)2=25
3-4x=5
4x=3-5
4x=-2
x=-2:4
x=-0,5
b)(2x-1/42)=16
2x-1/4=4
2x=4+1/4
2x=4,25
x=2,125
c) cái này x ở đâu vậy bn
d) (2x+5)2=0
2x+5=0
2x=0+5
2x=5
x=5:2
x=5/2
Nhớ k cho mk nha