tìm x:
I x-5 I = 7 - ( -3 )
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|x-1|+|x-2|+|x-4|=3(1)
TH1: x<1
Phương trình (1) sẽ trở thành:
1-x+2-x+4-x=3
=>7-3x=3
=>3x=4
=>\(x=\dfrac{4}{3}\left(loại\right)\)
TH2: 1<=x<2
Phương trình (1) sẽ trở thành:
x-1+2-x+4-x=3
=>-x+5=3
=>-x=-2
=>x=2(loại)
TH3: 2<=x<4
Phương trình (1) sẽ trở thành:
x-1+x-2+4-x=3
=>x+1=3
=>x=2(nhận)
TH4: x>=4
Phương trình (1) sẽ trở thành:
x-1+x-2+x-4=3
=>3x-7=3
=>3x=10
=>\(x=\dfrac{10}{3}\left(loại\right)\)
a) |x-3|+5=-7
|x+3|=-7-5
|x-3|=-12( vô lí)
Vậy không có giá trị nào của x thỏa mãn đề bài
b)||x-7|-6|=3
suy ra |x-7|-6=3 hoặc |x-7|-6 = -3
|x-7|=3+6 |x-7|=-3+6
|x-7|=9 |x-7|=3
suy ra x-7=9 hoặc x-7=-9 hoặc x-7=3 hoặc x-7=-3
TH1 ; x-7=9 TH2 : x-7=-9 TH3 : x-7=3 TH4 : x-7=-3
x=9+7 x= -9+7 x=3+7 x=-3+7
x=16 x=-2 x=10 x=4
Vậy x thuộc {10;-2;10;4}
c) |x-1|=5-x
suy ra x-1=5-x hoặc x-1=-(5-x)
x+x=1+5 x-1=-5+x
2x=6 x-x=1-5|(vô lí)
x=6:2
x=3
Vậy x=3
d) |1-x|+x+3=6
|1-x|+x=6-3
|1-x|+x=3
|1-x|=3-x
suy ra 1-x=3-x hoặc 1-x=-(3-x)
TH1 : 1-x=3-x TH2 : 1-x=(-3-x)
-x+x=-1+3 (vô lí ) 1-x=-3+x
-x-x=-1-3
-2x=-4
x= -4:(-2)
x=2
vậy x=2
c: Ta có: \(\left|x+\dfrac{5}{6}\right|:\dfrac{4}{5}=\dfrac{3}{8}\)
\(\Leftrightarrow\left|x+\dfrac{5}{6}\right|=\dfrac{3}{8}\cdot\dfrac{4}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{6}=\dfrac{3}{10}\\x+\dfrac{5}{6}=-\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-8}{15}\\x=-\dfrac{17}{15}\end{matrix}\right.\)
a) x + 1/3 = 3/4 b) x - 2/5 = 5/7
x = 3/4 - 1/3 x = 5/7 + 2/5
x = 5/12 x = 39/35
d) 4/7 - x = 1/3
x =4/7 - 1/3
x =5/21
mình làm đc rồi nha nhớ k mình à mà câu c mình ko hiểu mình ko có làm đc
a) 5.2² + (x + 3) = 5²
5.4 + x + 3 = 25
20 + x + 3 = 25
x + 23 = 25
x = 25 - 23
x = 2
b) 2³ + (x - 3²) = 5³ - 4³
8 + (x - 9) = 125 - 64
8 + x - 9 = 61
x - 1 = 61
x = 61 + 1
x = 62
c) 4.(x - 5) - 2³ = 2⁴.3
4x - 20 - 8 = 16.3
4x - 28 = 48
4x = 48 + 28
4x = 76
x = 76 : 4
x = 19
d) 5.(x + 7) - 10 = 2³.5
5x + 35 - 10 = 8.5
5x + 25 = 40
5x = 40 - 25
5x = 15
x = 15 : 5
x = 3
e) 7² - 7.(13 - x) = 14
49 - 91 + 7x = 14
7x - 42 = 14
7x = 14 + 42
7x = 56
x = 56 : 7
x = 8
a) \(5\cdot2^2+\left(x+3\right)=5^2\)
\(\Rightarrow x+3=5^2-5\cdot2^2\)
\(\Rightarrow x+3=25-5\cdot4\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=5-3\)
\(\Rightarrow x=2\)
b) \(2^3+\left(x-3^2\right)=5^3-4^3\)
\(\Rightarrow8+\left(x-9\right)=125-64\)
\(\Rightarrow8+x-9=61\)
\(\Rightarrow x-1=61\)
\(\Rightarrow x=61+1\)
\(\Rightarrow x=62\)
c) \(4\left(x-5\right)-2^3=2^4\cdot3\)
\(\Rightarrow4\left(x-5\right)=2^4\cdot3+2^3\)
\(\Rightarrow4\cdot\left(x-5\right)=16\cdot3+8\)
\(\Rightarrow4\cdot\left(x-5\right)=56\)
\(\Rightarrow x-5=56:4\)
\(\Rightarrow x-5=14\)
\(\Rightarrow x=19\)
d) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)=8\cdot5+10\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow5\left(x+7\right)=50\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
e) \(7^2-7\left(13-x\right)=14\)
\(\Rightarrow7\left(13-x\right)=7^2-14\)
\(\Rightarrow7\left(13-x\right)=49-14\)
\(\Rightarrow7\left(13-x\right)=35\)
\(\Rightarrow13-x=5\)
\(\Rightarrow x=13-5\)
\(\Rightarrow x=8\)
f) \(5x-5^2=10\)
\(\Rightarrow5x=10+5^2\)
\(\Rightarrow5x=10+25\)
\(\Rightarrow5x=35\)
\(\Rightarrow x=\dfrac{35}{5}\)
\(\Rightarrow x=7\)
g) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x=3^4+2\cdot3^2\)
\(\Rightarrow9x=81+2\cdot9\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
h) \(10x+2^2\cdot5=10^2\)
\(\Rightarrow10x=10^2-2^2\cdot5\)
\(\Rightarrow10x=100-4\cdot5\)
\(\Rightarrow10x=80\)
\(\Rightarrow x=\dfrac{80}{10}\)
\(\Rightarrow x=8\)
i) \(125-5\left(4+x\right)=15\)
\(\Rightarrow5\left(4+x\right)=125-5\)
\(\Rightarrow5\left(4+x\right)=120\)
\(\Rightarrow4+x=\dfrac{120}{5}\)
\(\Rightarrow4+x=24\)
\(\Rightarrow x=24-4\)
\(\Rightarrow x=20\)
j) \(2^6+\left(5+x\right)=3^4\)
\(\Rightarrow5+x=3^4-2^6\)
\(\Rightarrow5+x=81-64\)
\(\Rightarrow5+x=17\)
\(\Rightarrow x=17-5\)
\(\Rightarrow x=12\)
a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)
=>\(\left|2x+1\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: (2x-3)2=36
=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: \(7^{x+2}+2\cdot7^x=357\)
=>\(7^x\cdot49+7^x\cdot2=357\)
=>\(7^x=7\)
=>x=1
a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
\(---\)
b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)
\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)
\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
\(---\)
c) \(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(---\)
d) \(7^{x+2}+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)
\(\Rightarrow7^x\cdot\left(49+2\right)=357\)
\(\Rightarrow7^x\cdot51=357\)
\(\Rightarrow7^x=357:51\)
\(\Rightarrow7^x=7\)
\(\Rightarrow x=1\)
a/ (ghi lại cái đề)
=>+ 3x-7=2
3x=2+7=9
x=3
+ 3x-7=-2
3x=-2+7=5
x=\(\frac{5}{3}\)
b/ (5x-10)2=100
=> +5x-10=10
5x=10+10=20
x=4
+ 5x-10=-10
5x=-10+10=0
x=0
\(\left(|x+3|-5\right)\left(x^2+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}|x+3|-5=0\\x^2+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}|x+3|=5\\x^2=-7\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=5\\x+3=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-8\end{cases}}}\)
Loại \(x^2=-7\)vì \(x^2\ge0\forall x\)
Vậy x=2; x=-8
\(\left(|x+3|-5\right)\left(x^2+7\right)=0\)
\(TH1:|x+3|-5=0\)
\(|x-3|=5\)
\(\Rightarrow\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\)
\(TH2:x^2+7=0\)
\(x^2=-7\)
\(\Rightarrow x=\varnothing\)
các bạn giúp mình với
=>|x-5|=10
Ta có:|x-5|=10=>x-5€{10;-10}
TH1:x-5=10
=>x=15
TH2:x-5=-10
=>x=-5
Vậy x €{-5;15}