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a) \(\left(x^2+1\right)\left(2x-4\right)>0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x^2+1\right)>0\\\left(2x-4\right)>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2>-1\\2x>4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x>\sqrt{-1}\\x>2\end{cases}}}\)
Vậy \(x>2\)
b)\(\left(5x-15\right)\left(x^2+1\right)< 0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(5x-15\right)< 0\\x^2+1< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x< 15\\x^2< -1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< \sqrt{-1}\end{cases}}}\)
Vậy \(x< \sqrt{-1}\)
\(2x-8x^2=0\Rightarrow2x\left(1-4x\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\1-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)
\(x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Cn lại lm tương tự nha e!
=.= hok tốt!!
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
Bài này giống tìm nghiệm quá :
1) \(5\left(x-7\right)=0\)
\(\left(x-7\right)=0\div5\)
\(\left(x-7\right)=0\)
\(x=0+7\)
\(x=7\)
2) \(25\left(x-4\right)=0\)
\(\left(x-4\right)=0\div25\)
\(\left(x-4\right)=0\)
\(x=0+4\)
\(x=4\)
3) \(34\left(2x-6\right)=0\)
\(\left(2x-6\right)=0\div34\)
\(\left(2x-6\right)=0\)
\(2x=0+6\)
\(2x=6\)
\(x=6\div2\)
\(x=3\)
4) \(2007\left(3x-12\right)=0\)
\(\left(3x-12\right)=0\div2007\)
\(\left(3x-12\right)=0\)
\(3x=0+12\)
\(3x=12\)
\(x=12\div3\)
\(x=4\)
5) \(47\left(5x-15\right)=0\)
\(\left(5x-15\right)=0\div47\)
\(\left(5x-15\right)=0\)
\(5x=0+15\)
\(5x=15\)
\(x=15\div5\)
\(x=3\)
6) \(13\left(4x-24\right)=0\)
\(\left(4x-24\right)=0\div13\)
\(\left(4x-24\right)=0\)
\(4x=0+24\)
\(4x=24\)
\(x=24\div4\)
\(x=6\)
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
a) x = { -5 ; -2 }
b) x = { -1 ; 0 ; 1 ;2 ; 3 ; .........}
c) x = { -4 ; 3 }
d) x = { -3 ; 2 }
e) x = { 1 ; 2 ; 3 ; 4 ; .............}
Tớ ko chắn chắn cho lắm !!!!! Các bạn thấy đúg thì Tick nha !!!!!
\(\left(x-5\right)\left(x+4\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=0+5\\x=0-4\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
\(\Rightarrow x=5\) hoặc \(x=-4\)
\(\left|-15\right|-\left|x\right|=\left|-12\right|\)
\(15-\left|x\right|=12\)
\(\left|x\right|=15-12\)
\(\left|x\right|=3\)
\(\Rightarrow x=\pm3\)
5x+9=2x+15<=>5x-2x=15-9<=>3x=6<=>x=3
(x-5).(x+4)=0
Th1x-5=0<=>x=5
Th2x+4=0<=>x=-4
Vậy x=0;-4
2.(x-3)-4(x+4)=3.(-7)+5<=>2x-6-4x-16=-21+5<=>-2x-22=-16<=>-2x=-16+22
<=>-2x=6<=>2x=-6<=>x=-3
/-15/-/x/=/-12/
Th1:x<0=>/-15/-/x/=/-12/<=>15-(-x)=12<=>15+x=12<=>x=27
Th2:x> hoặc= 0=>/-15/-/x/=/-12/<=>15-x=12<=>15-12=x<=>x=3
\(\orbr{\begin{cases}x-4=0\\5x-15=0\end{cases}}\)
\(\orbr{\begin{cases}x=4\\x=3\end{cases}}\)
xin tiick
(5𝑥−15)(𝑥−4)=0
(5x-15)(x-4)}}=0(5x−15)(x−4)=0
5𝑥(𝑥−4)−15(𝑥−4)=0
Đáp án
𝑥=4𝑥=3