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h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)
=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)
d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
=> 2x-1 = \(\dfrac{-2}{3}\)
=> x= \(\dfrac{1}{6}\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
2) \(\dfrac{x}{y}=\left(\dfrac{x}{y}\right)^2\)
\(\Rightarrow\left(\dfrac{x}{y}\right)^2-\dfrac{x}{y}=0\)
\(\Rightarrow\dfrac{x}{y}\left(\dfrac{x}{y}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{y}=0\Rightarrow x=0;y\in R\\\dfrac{x}{y}-1=0\Rightarrow\dfrac{x}{y}=1\Rightarrow x=y\end{matrix}\right.\)
3) \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}.2^5+2^{15}.1=2^{15}.33⋮33\rightarrowđpcm\)
4)\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)
\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\\\left(x-4-y\right)^{200}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-12+y\right)^{200}+\left(x-y-4\right)^{200}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-y-4\right)^{200}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-12+y=0\Rightarrow x+y=12\\x-y-4=0\Rightarrow x-y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)+\left(x-y\right)=12+4\Rightarrow x+y+x-y=16\Rightarrow2x=16\Rightarrow x=8\\y=8-4=4\end{matrix}\right.\)
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
a: Đặt A=0
=>-2/3x=5/9
hay x=-5/6
b: Đặt B(x)=0
=>(x-2/5)(x+2/5)=0
=>x=2/5 hoặc x=-2/5
c: Đặt C(X)=0
\(\Leftrightarrow x^3\cdot\dfrac{1}{2}=-\dfrac{4}{27}\)
\(\Leftrightarrow x^3=-\dfrac{8}{27}\)
hay x=-2/3
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
a/ \(\dfrac{x}{9}=\dfrac{16}{x}\)
\(\Leftrightarrow x^2=9.16\)
\(\Leftrightarrow x^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right.\)
Vậy ...
b/ \(x^3+27=0\)
\(\Leftrightarrow x^3=-27\)
\(\Leftrightarrow x^3=\left(-3\right)^3\)
\(\Leftrightarrow x=-3\)
Vậy ...
c/ \(\left|x\left(x^2-\dfrac{5}{4}\right)=x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{5}{4}\right)=x\\x\left(x^2-\dfrac{5}{4}\right)=-x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{5}{4}x=x\\x^3-\dfrac{5}{4}x=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^3-\left(\dfrac{5}{4}x+x\right)=0\\x^3-\left(\dfrac{5}{4}x-x\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{9}{4}x=0\\x^3-\dfrac{1}{4}x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{9}{4}\right)=0\\x\left(x^2-\dfrac{1}{4}\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{4}=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
c/ Với mọi x ta có :
\(\left|x-5\right|=\left|5-x\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|x-5\right|=\left|x+3\right|+\left|5-x\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge\left|\left(x+3\right)+\left(5-x\right)\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge\left|8\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge8\)
Dấu "=" xảy ra khi :
\(\left(x+3\right)\left(5-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3\\5\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3\\5\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3\le x\le5\\x\in\varnothing\end{matrix}\right.\)
Vậy ...