![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$(x-15).27=0$
$x-15=0:27=0$
$x=15+0=15$
b.
$23(42-x)=0$
$42-x=0$
$x=42$
c.
$(9x+2).3=60$
$9x+2=60:3=20$
$9x=18$
$x=2$
d.
$71+(26-3x):5=75$
$(26-3x):5=75-71=4$
$26-3x=4.5=20$
$3x=26-20=6$
$x=6:2=3$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\) Vì \(x,y\in Z\) nên \(\left(3x+2\right):3R2;R1\)
Mà \(\left(3x+2\right)\left(y-8\right)=12\) nên \(3x+2\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Do đó \(3x+2\in\left\{-4;-1;2\right\}\)
\(\Rightarrow x\in\left\{-2;-1;0\right\}\)
Với \(x=-2\Rightarrow\left(-4\right)\left(y-8\right)=12\Rightarrow y-8=-3\Rightarrow y=5\)
Với \(x=-1\Rightarrow\left(-3\right)\left(y-8\right)=12\Rightarrow y-8=-4\Rightarrow y=4\)
Với \(x=0\Rightarrow2\left(y-8\right)=12\Rightarrow y-8=6\Rightarrow y=14\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-2;5\right);\left(-1;4\right);\left(0;14\right)\)
\(b,\) Vì \(x,y\in Z\) nên \(\left(5x-4\right):5R1;R4\)
Mà \(\left(5x-4\right)\left(y+3\right)=-18\)
\(\Rightarrow5x-4\inƯ\left(-18\right)=\left\{-18;-9;-6;-3;-2;-1;1;2;3;6;9;18\right\}\\ \Rightarrow5x-4\in\left\{-9;1;6\right\}\\ \Rightarrow x\in\left\{-1;1;2\right\}\)
Với \(x=-1\Rightarrow-9\left(y+3\right)=-18\Rightarrow y+3=2\Rightarrow y=-1\)
Với \(x=1\Rightarrow y+3=18\Rightarrow y=15\)
Với \(x=2\Rightarrow6\left(y+3\right)=18\Rightarrow y+3=3\Rightarrow y=0\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-1;-1\right);\left(1;15\right);\left(2;0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ủa cho em hỏi ý a,b,c đâu ạ . chứ chị giải kiểu vậy em hong có hỉu , mong chị trả lời em ạ
![](https://rs.olm.vn/images/avt/0.png?1311)
b: =>x(8-7)=-33
=>x=-33
c: =>-12x+60+21-7x=5
=>-19x=-76
hay x=4
d: =>-2x-2-x+5+2x=0
=>3-x=0
hay x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Sửa đề: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{2}{-z}=\dfrac{-t}{-9}\)
=>\(\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{-2}{z}=\dfrac{t}{9}=-2\)
=>\(x=-2\cdot5=-10;y=-2\cdot\left(-3\right)=6;z=\dfrac{-2}{-2}=1;t=9\cdot\left(-2\right)=-18\)
b: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)
=>\(\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)
=>\(\left\{{}\begin{matrix}x=4\cdot3=12\\y^2=\dfrac{4}{4}=1\\z^3=-2\cdot4=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y\in\left\{1;-1\right\}\\z=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>3^x=3^4*3=3^5
=>x=5
b: =>\(2^{x+1}=2^5\)
=>x+1=5
=>x=4
c: \(\Leftrightarrow3^{x+2-3}=3\)
=>x-1=1
=>x=2
d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)
=>x=4 hoặc x=-4
e: (2x-1)^4=81
=>2x-1=3 hoặc 2x-1=-3
=>2x=4 hoặc 2x=-2
=>x=-1 hoặc x=2
f: (2x-6)^4=0
=>2x-6=0
=>x-3=0
=>x=3
a) \(3^x=81\cdot3\)
\(\Rightarrow3^x=3^4\cdot3\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
b) \(2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
c) \(3^{x+2}:27=3\)
\(\Rightarrow3^{x+2}:3^3=3\)
\(\Rightarrow3^{x+2-3}=3\)
\(\Rightarrow3^{x-1}=3\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
d) \(2x^2=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
e) \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
f) \(\left(2x-6\right)^4=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=6:2\)
\(\Rightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,-\dfrac{13}{20}+x=\dfrac{-11}{15}\\ \Rightarrow x=\dfrac{-11}{15}+\dfrac{13}{20}\\ \Rightarrow x=-\dfrac{1}{12}\\ b,\left(x-3,5\right):3\dfrac{1}{2}-2,5=-1\dfrac{3}{4}\\ \Rightarrow\left(x-\dfrac{7}{2}\right):\dfrac{7}{2}-\dfrac{5}{2}=\dfrac{-7}{4}\\ \Rightarrow\left(x-\dfrac{7}{2}\right):\dfrac{7}{2}=\dfrac{3}{4}\\ \Rightarrow x-\dfrac{7}{2}=\dfrac{21}{8}\\ \Rightarrow x=\dfrac{49}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x-43=\left(35-x\right)-48\)
=>\(x-43=35-x-48\)
=>\(x-43=-x-13\)
=>\(x+x=-13+43\)
=>2x=30
=>x=30/2=15
b: \(305-x+14=48+\left(x-23\right)\)
=>\(319-x=48+x-23=25+x\)
=>\(x+25=319-x\)
=>\(x+x=319-25\)
=>\(2x=294\)
=>\(x=\dfrac{294}{2}=147\)
c: \(-\left(-x-6+85\right)=\left(x+51\right)-54\)
=>\(-\left(-x+79\right)=x+51-54\)
=>x-79=x-3
=>-79=-3(vô lý)
=>\(x\in\varnothing\)
d: \(-\left(35-x\right)-\left(37-x\right)=33-x\)
=>\(-35+x-37+x=33-x\)
=>2x-72=-x+33
=>\(2x+x=33+72\)
=>3x=105
=>\(x=\dfrac{105}{3}=35\)
a) \(\left(x-2\right)\left(x+2\right)>0\)
\(\Leftrightarrow x^2-4\ge0\)
\(\Leftrightarrow x\notin\left\{-1;0;1\right\}\)
\(\left(x-2\right)\left(x+2\right)>0\)
\(\Rightarrow\hept{\begin{cases}x-2\\x+2\end{cases}}\)cùng dấu
Trường hợp 1 : \(x-2\)và \(x+2\)cùng dương
\(\Rightarrow\hept{\begin{cases}x-2>0\\x+2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0+2\\x>0-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>2\\x>-2\end{cases}}\left(\text{vô lí}\right)\)
Nên ta loại trường hợp 1
Trường hợp 2 : \(x-2\)và \(x+2\)cùng âm
\(\Rightarrow\hept{\begin{cases}x-2< 0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 0+2\\x< 0-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2\\x< -2\end{cases}}\left(\text{}\text{vô lí}\right)\)
Nên ta loại trường hợp 2
Trường hợp 3 : \(x-2< x+2\)luôn đúng
\(\Rightarrow x\ge2\)
\(\left(x-2\right)\left(x+3\right)=15\)
Lập bảng là ra