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N
31 tháng 7 2017
1.
a) \(\left|5-2x\right|:3-2,6=0\)
\(\left|5-2x\right|=7,8\)
\(\Rightarrow\left[{}\begin{matrix}5-2x=7,8\\5-2x=-7,8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1,4\\x=6,4\end{matrix}\right.\)
Vậy ....
b) \(\left|2x-1\right|.5-7=0\)
\(\left|2x-1\right|=1.4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=1,4\\2x-1=-1,4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,2\\x=-0,2\end{matrix}\right.\)
Vậy...
c) \(\left|x+1\right|+\left|x-2\right|=1\)
* Nếu \(x< -1\) => \(\left\{{}\begin{matrix}x+1< 0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left|x+1\right|=-x-1\\\left|x-2\right|=2-x\end{matrix}\right.\)
Khi đó \(-x-1+2-x=1\)
\(\Rightarrow x=0\) ( loại vì x > -1)
* Nếu \(-1\le x< 2\)\(\Rightarrow\left\{{}\begin{matrix}x+1\ge0\\x-2< 0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x-2\right|=2-x\end{matrix}\right.\)
Khi đó \(x+1+2-x=1\)
\(\Rightarrow3=1\) (Vô lí)
* Nếu \(x\ge2\Rightarrow\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x-2\right|=x-2\end{matrix}\right.\)
Khi đó \(x+1+x-2=1\)
\(x=1\)(loại)
Vậy ...
tik mik nha !!!
27 tháng 8 2017
a) \(\left(x-1\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)
b) \(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x=5\)
c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x\in\varnothing\)
12 tháng 1 2020
a) (2x-1)(x+2)=0
<=> 2x-1=0
Hoặc x+2=0
<=>2x=1
Hoặc x=-2
<=>x=1/2
Hoặcx=-2
Xl nha mk ko bt làm mấy câu kia
AT
20 tháng 6 2017
a/dễ --> tự lm
b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...............
c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)
TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)
TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)
Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề
d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)
TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)
TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)
Vậy...................
25 tháng 7 2018
a, \(\left(x-3\right)\left(2x+5\right)>0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\2x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\2x+5< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< -\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -\dfrac{5}{2}\end{matrix}\right.\)
b,\(\left(1-4x\right)\left(x-2\right)< 0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-4x>0\\x-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-4x< 0\\x-2>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{1}{4}\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{1}{4}\\x>2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 2\\x>2\end{matrix}\right.\)
25 tháng 7 2018
c, \(\dfrac{-3}{x+2}< 0\Leftrightarrow x+2>0\Leftrightarrow x>-2\)
15 tháng 7 2022
a: \(\Leftrightarrow25\left(x+1\right)^4-25\left(x+1\right)^2-\left(x+1\right)^2+1=0\)
\(\Leftrightarrow\left[\left(x+1\right)^2-1\right]\left[25\left(x+1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x+2\right)\cdot x\cdot\left(5x+4\right)\left(5x+6\right)=0\)
hay \(x\in\left\{0;-2;-\dfrac{4}{5};-\dfrac{6}{5}\right\}\)
b: \(x^2+x-1=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-1\right)=5\)
Do đó: PT có 2 nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{5}}{2}\\x_2=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)
d: \(\Leftrightarrow4x^2-4x+1-5\left(2x-1\right)-6=0\)
\(\Leftrightarrow\left(2x-1\right)^2-5\left(2x-1\right)-6=0\)
=>(2x-1-6)(2x-1+1)=0
=>(2x-7)2x=0
=>x=0 hoặc x=7/2
AT
7 tháng 7 2018
a/ |2x - 3| + |y - 2| = 0
Vì: \(\left\{{}\begin{matrix}\left|2x-3\right|\ge0\forall x\\\left|y-2\right|\ge0\forall y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}2x-3=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=2\end{matrix}\right.\)
b/ |3x - 4| + |x - y| = 0
Vì: \(\left\{{}\begin{matrix}\left|3x-4\right|\ge0\forall x\\\left|x-y\right|\ge0\forall x;y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3x-4=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\x=y=\dfrac{4}{3}\end{matrix}\right.\)
Vậy x = y = 4/3
c/ \(\left|2x+y-1\right|+\left|2y-3\right|=0\)
Vì: \(\left\{{}\begin{matrix}\left|2x+y-1\right|\ge0\forall x;y\\\left|2y-3\right|\ge0\forall y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}2x+y-1=0\\2y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-1=-y\\y=\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=-\dfrac{3}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=\dfrac{3}{2}\end{matrix}\right.\)
Vậy..........
d/ \(\left|x+y-5\right|+\left|2x-y+8\right|=0\)
Vì: \(\left\{{}\begin{matrix}\left|x+y-5\right|\ge0\\\left|2x-y+8\right|\ge0\end{matrix}\right.\)∀x;y
=> \(\left\{{}\begin{matrix}x+y-5=0\\2x-y+8=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\2x-y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5-y\\2\left(5-y\right)-y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5-y\\10-2y-y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5-y\\-3y=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5-y\\y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5-6=-1\\y=6\end{matrix}\right.\)
Vậy x = -1; y = 6
LV
28 tháng 9 2018
a/ |2x - 3| + |y - 2| = 0
Vì: {|2x−3|≥0∀x|y−2|≥0∀y{|2x−3|≥0∀x|y−2|≥0∀y
=> {2x−3=0y−2=0⇒⎧⎨⎩x=32y=2{2x−3=0y−2=0⇒{x=32y=2
b/ |3x - 4| + |x - y| = 0
Vì: {|3x−4|≥0∀x|x−y|≥0∀x;y{|3x−4|≥0∀x|x−y|≥0∀x;y
=> {3x−4=0x−y=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=43x=y=43{3x−4=0x−y=0⇔{x=43x=y=43
Vậy x = y = 4/3
c/ |2x+y−1|+|2y−3|=0|2x+y−1|+|2y−3|=0
Vì: {|2x+y−1|≥0∀x;y|2y−3|≥0∀y{|2x+y−1|≥0∀x;y|2y−3|≥0∀y
=> {2x+y−1=02y−3=0⇔⎧⎨⎩2x−1=−yy=32{2x+y−1=02y−3=0⇔{2x−1=−yy=32
⇔⎧⎪ ⎪⎨⎪ ⎪⎩2x−1=−32y=32⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=−14y=32⇔{2x−1=−32y=32⇔{x=−14y=32
Vậy..........
d/ |x+y−5|+|2x−y+8|=0|x+y−5|+|2x−y+8|=0
Vì: {|x+y−5|≥0|2x−y+8|≥0{|x+y−5|≥0|2x−y+8|≥0∀x;y
=> {x+y−5=02x−y+8=0{x+y−5=02x−y+8=0⇔{x+y=52x−y=−8⇔{x+y=52x−y=−8
⇔{x=5−y2(5−y)−y=−8⇔{x=5−y2(5−y)−y=−8
⇔{x=5−y10−2y−y=−8⇔{x=5−y10−2y−y=−8
⇔{x=5−y−3y=−18⇔{x=5−yy=6⇔{x=5−6=−1y=6⇔{x=5−y−3y=−18⇔{x=5−yy=6⇔{x=5−6=−1y=6
Vậy x = -1; y = 6
CHÚC BẠN HỌC TỐT
11 tháng 6 2018
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
TT
11 tháng 6 2018
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: \(S=\left\{12;-\dfrac{12}{7}\right\}\)
25 tháng 7 2017
\(\left(x-1\right)\left(x+5\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\end{matrix}\right.\)
\(\left(x-1\right)\left(x+5\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-5< x< 1\)
câu dễ tự làm
\(\Rightarrow x>-5;x< -5\)
6 tháng 2 2016
a/ x2+5x=0
=> x2=5x=0
=> x=0
b/ 3(2x+3)(3x-5)<0
=> 2x+3 và 3x-5 phải khác dấu
x=0
câu này mk chỉ bít kết quả thui thông cảm nha
TH1 : x + 5 = 0
x = 0 - 5
x = -5
TH2 : 1 - 2x = 0
2x = 1-0
2x = 1
x = 1 : 2
x = 1/2
Vậy x thuộc { -5 ; 1/2 }
Ta có : | x+5| \(\ge\) 0 ; |1-2x| \(\ge\) 0
=> |x+5 | + |1-2x | \(\ge\) 0
mà |x+5 | + |1-2x | = 0
=> \(\hept{\begin{cases}x+5=0\\1-2x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\x=\frac{1}{2}\end{cases}}}\)
Vây x =-5 ; x=1/2