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`|2x+1|-3=x+4`
`<=>|2x+1|=x+4+3=x+7(x>=-7)`
`**2x+1=x+7`
`<=>x=7-1=6(tm)`
`**2x+1=-x-7`
`<=>3x=-6`
`<=>x=-2(tm)`
`|3x-5|=1-3x(x<=1/3)`
`**3x-5=1-3x`
`<=>6x=6`
`<=>x=1(l)`
`**3x-5=3x-1`
`<=>-5=-1` vô lý
`|2x+2|+|x-1|=10`
Nếu `x>=1`
`pt<=>2x+2+x-1=10`
`<=>3x+1=10`
`<=>3x=9`
`<=>x=3(tm)`
Nếu `x<=-1`
`pt<=>-2x-2+1-x=10`
`<=>-1-3x=10`
`<=>-11=3x`
`<=>x=-11/3(tm)`
Nếu `-1<=x<=1`
`pt<=>2x+2+1-x=10`
`<=>x+3=10`
`<=>x=7(l)`
Vậy `S={3,-11/3}`
`a,`\(2^x -15= 2^4+1\)
`-> 2^x-15=17`
`-> 2^x=17+15`
`-> 2^x=32`
`-> 2^x=2^5`
`-> x=5`
`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?
`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)
`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)
`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
Mà `1/65+1/64-1/63-1/62 \ne 0`
`-> x+66=0`
`-> x=-66`
a: =>2^x=2^4+16=32
=>x=5
b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)
=>x+66=0
=>x=-66
\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)
\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)
\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)
\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)
\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)
\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)
\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)
a: \(\Leftrightarrow-x^2-3x+x+3+x^2-6x=11\)
=>-8x+3=11
=>-8x=8
hay x=-1
b: \(\Leftrightarrow3x^2-15x+x-5-3x^2+3x=5\)
=>-11x=10
hay x=-10/11
\(\dfrac{x}{y}=\dfrac{-3}{4}\)
⇒\(\dfrac{x}{-3}=\dfrac{y}{4}\)
⇒\(\dfrac{2x}{-6}=\dfrac{3y}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{-6}=\dfrac{3y}{12}=\dfrac{3y-2x}{12-\left(-6\right)}=\dfrac{36}{18}=2\)
⇒\(\left\{{}\begin{matrix}x=2.-3=-6\\y=2.4=8\end{matrix}\right.\)
a, /3x-2/-x>1
3x-2-x>1
2x-2>1
2x>3
x>3/2
b,/2x+3/<hoac =5
2x+3<hoac =5
2x</=2
x</=1
cho nhiuf nha banj!
cho mih nha