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\(4^x+4^{x+3}=1040\)
\(4^x.1+4^x.4^3=1040\)
\(4^x.\left(1+4^3\right)=1040\)
\(4^x.65=1040\)
\(4^x=1040:65\)
\(4^x=16\)
\(4^x=4^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
1) (4x−7)2−5×|7−4x|=0
Có (4x-7)2 \(\ge0\) với mọi x
|7−4x| \(\ge0\) với mọi x
<=> 5|7−4x| \(\ge0\) với mọi x
Để (4x−7)2−5×|7−4x|=0 thì \(\left\{{}\begin{matrix}\left(4x-7\right)^2=0\\5|7-4x|=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}4x-7=0\\7-4x=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}4x=7\\4x=7\end{matrix}\right.\)<=>\(x=\dfrac{7}{4}\)
Vậy \(x=\dfrac{7}{4}\)
2) \(4^{x-2}+4^{x+1}=1040\)
<=> \(4^{x+1}.4^{-3}+4^{x+1}=1040\)
<=> \(4^{x+1}\left(4^{-3}+1\right)=1040\)
<=> \(4^{x+1}.\dfrac{65}{64}=1040\)
<=> \(4^{x+1}=1024=4^5\)
=> x+1=5 <=> x=4
Vậy x=4
a ) 4x+2 +4x+1 = 1040
4x.42+4x.4=1040
4x.(42+4)=1040
4x.(16+4)=1040
4x.20=1040
4x=1040:20
4x=52
Vô lí vì 52 ko chuyển thành 4 mũ mấy đc
Vậy \(x\in\varnothing\)
\(b)\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)
Vô lí vì \(\frac{3}{4}\)ko chuyển thành đc mũ 2
Vậy \(x\in\varnothing\)
Mình sẽ giúp bạn làm câu còn lại :)))
\(\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)
\(\Leftrightarrow x-\sqrt{3}=\pm\sqrt{\frac{3}{4}}\)
\(\Leftrightarrow x-\sqrt{3}=\pm\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{3}=\frac{\sqrt{3}}{2}\\x-\sqrt{3}=-\frac{\sqrt{3}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3\sqrt{3}}{2}\\x=\frac{\sqrt{3}}{2}\end{cases}}\)
a, \(\left[x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\right]x^2-1\)
\(=\left[x\left(x^2-16\right)-\left(x^2+1\right)\right]x^2-1\)
\(=\left[x^3-16x-x^2-1\right]x^2-1\)
\(=x^5-16x^3-x^4-x^2-1\)
b, \(\left(y-3\right)y+3y^2+9-y^2+2\left(y^2-2\right)\)
\(=y^2-3y+3y^2+9-y^2+2y^2-4\)
\(=5y^2-3y+5\)
c, \(\left(x+y\right)\left(x^2x^2-xy+y^2\right)\)
\(=x^5-x^2y+xy^2+x^4y-xy^2+y^3\)
d, \(\left(\dfrac{1}{2}xy+\dfrac{3}{4}y\right).\dfrac{1}{2}xy-\dfrac{3}{4}y\)
\(=\dfrac{1}{4}x^2y^2+\dfrac{3}{8}xy^2-\dfrac{3}{4}y\)
\(=\dfrac{1}{4}y.\left(x^2y+\dfrac{3}{2}xy-3\right)\)
Chúc bạn học tốt!!!
a) \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=21+25\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
Vậy \(x=23\)
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Rightarrow\left(x-1\right).\left(x+1\right)=7.9\)
\(\Rightarrow\left(x-1\right)x-\left(x+1\right)=7.9\)
\(\Rightarrow x^2-x-x-1=63\)
\(\Rightarrow x^2-1=63\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
Vậy \(x=8\) hoặc \(x=-8\)
c) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10\)
+) \(x+4=10\Rightarrow x=6\)
+) \(x+4=-10\Rightarrow x=-16\)
Vậy \(x\in\left\{6;-16\right\}\)
a/ \(\dfrac{x+1}{2}=\dfrac{2x+3}{5}\)
\(\Leftrightarrow5\left(x+1\right)=2\left(2x+3\right)\)
\(\Leftrightarrow5x+5=4x+6\)
\(\Leftrightarrow5x-4x=6-5\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ...
b/ \(\left|x-1\right|+3\left|y+1\right|+\left|z+2\right|=0\)
Mà với \(\forall x;y;z\) ta có :
\(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\3\left|y+1\right|\ge0\\\left|z+2\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\3\left|y+1\right|=0\\\left|z+2\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\\z+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=-2\end{matrix}\right.\)
Vậy ...
c/ \(\dfrac{x-2}{4}=\dfrac{5-3x}{4}\)
\(\Leftrightarrow x-2=5-3x\)
\(\Rightarrow x+3x=5+2\)
\(\Leftrightarrow4x=7\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
Vậy ......
d/ \(\dfrac{x+2}{4}=\dfrac{4}{x+2}\)
\(\Leftrightarrow\left(x+2\right)\left(x+2\right)=16\)
\(\Leftrightarrow\left(x+2\right)^2=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy ...
e/ \(\dfrac{x-1}{5}=\dfrac{-20}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=-100\)
\(\Leftrightarrow\left(x-1\right)^2=-100\)
Lại có : \(\left(x-1\right)^2\ge0\)
\(\Leftrightarrow\) k tồn tại x
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
4^x . 4^2 + 4^x . 4 =1040
4^x . ( 16+4) =1040
4^x . 20=1040
4^x=1040:20=52
=> x thuộc tập hợp rỗng
Thanks