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a, Ta có:
\(\left|2x+4\right|+\left|4x+8\right|\ge0\)
Để \(\left|2x+4\right|+\left|4x+8\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|2x+4\right|=0\\\left|4x+8\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=-2\end{matrix}\right.\Rightarrow x=-2\)
Vậy...........
b, Ta có:
\(\left|x-5\right|+\left|x-7\right|\ge0\)
Để \(\left|x-5\right|+\left|x-7\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x-5\right|=0\\\left|x-7\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=7\end{matrix}\right.\Rightarrow x\in\varnothing\)
Vậy...........
c,\(\left|x+8\right|-\left|2x+2\right|=0\)
\(\Rightarrow\left|x+8\right|=\left|2x+2\right|\)
\(\Rightarrow\left\{{}\begin{matrix}x+8=2x+2\\x+8=-2x-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}-x=-6\\3x=-10\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy...........
Chúc bạn học tốt!!!
e) \(\left(x-3\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left(x-3\right)=0\) ( \(x^2+1>0\forall x\))
\(\Rightarrow x=3\)
đ) \(4.8^2=2^x\)
\(2^2.\left(2^3\right)^2=2^x\)
\(2^2.2^6=2^x\)
\(2^8=2^x\)
\(\Rightarrow x=8\)
d) \(\left|x+3\right|=8\)
\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)
mấy câu trên dễ rồi tự làm em nhé
a) \(2^{4x+1}-8^{x+2}=0\)\(\Leftrightarrow2^{4x+1}-2^{3\left(x+2\right)}=0\)
\(\Leftrightarrow2^{4x+1}-2^{3x+6}=0\)\(\Leftrightarrow2^{4x+1}=2^{3x+6}\)
\(\Leftrightarrow4x+1=3x+6\)\(\Leftrightarrow4x-3x=6-1\)\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(3^2.9^{2x}=27^{x+3}\)\(\Leftrightarrow3^2.3^{2.2x}=3^{3\left(x+3\right)}\)\(\Leftrightarrow3^2.3^{4x}=3^{3x+9}\)
\(\Leftrightarrow3^{2+4x}=3^{3x+9}\)\(\Leftrightarrow2+4x=3x+9\)\(\Leftrightarrow4x-3x=9-2\)\(\Leftrightarrow x=7\)
Vậy \(x=7\)
c) \(8^{2x}.64^2=16^{x+4}\)\(\Leftrightarrow2^{3.2x}.2^{6.2}=2^{4\left(x+4\right)}\)\(\Leftrightarrow2^{6x}.2^{12}=2^{4\left(x+4\right)}\)
\(\Leftrightarrow2^{6x+12}=2^{4x+16}\)\(\Leftrightarrow6x+12=4x+16\)\(\Leftrightarrow6x-4x=16-12\)
\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
a) \(\frac{2}{5}:\left(2x+\frac{3}{4}\right)=-\frac{7}{10}\)
=> \(2x+\frac{3}{4}=-\frac{7}{10}:\frac{2}{5}\)
=> \(2x+\frac{3}{4}=-\frac{7}{4}\)
=> \(2x=\frac{-7}{4}-\frac{3}{4}\)
=> \(2x=-\frac{5}{2}\)
=> \(x=\frac{-5}{2}:2\)
=> \(x=\frac{-5}{4}\)
b) \(\frac{x+1}{3}=\frac{2-x}{2}\)
\(\Rightarrow2\left(x+1\right)=3\left(2-x\right)\)
\(\Rightarrow2x+2=6-3x\)
\(\Rightarrow2x-3x=6-2\)
\(\Rightarrow-x=4\)
\(\Rightarrow x=4\)
c) \(\left|x-\frac{3}{5}\right|.\frac{1}{2}-\frac{1}{5}=0\)
\(\Rightarrow\left|x-\frac{3}{5}\right|.\frac{1}{2}=\frac{1}{5}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{1}{5}:\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}=\frac{2}{5}\\x-\frac{3}{5}=-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}+\frac{2}{5}\\x=\frac{3}{5}+-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
d) \(x^2-4x=0\)
Ta có : \(x^2-4x=0\)
\(\Rightarrow xx-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=0+4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Bài 2:
a) \(\left(x-3\right)^3+27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0-27\)
\(\Leftrightarrow\left(x-3\right)^3=-27\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=\left(-3\right)+3\)
\(\Leftrightarrow x=0\)
b) \(-125-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3=-125-0\)
\(\Leftrightarrow\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=\left(-5\right)-1\)
\(\Leftrightarrow x=-6\)
c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}:2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d) \(2^x+2^{x+1}=24\)
\(\Leftrightarrow2^x+2^x.2=24\)
\(\Leftrightarrow2^x\left(1+2\right)=24\)
\(\Leftrightarrow2^x.3=24\)
\(\Leftrightarrow2^x=24:3\)
\(\Leftrightarrow2^x=8\)
\(\Leftrightarrow2^x=2^3\)
\(\Rightarrow x=3\)
e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
g) \(\left|x-3\right|+2x=10\)
\(\Leftrightarrow\left|x-3\right|=10-2x\)
\(\Leftrightarrow\left|x-3\right|=2.5-2x\)
\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)
(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)
Bài 1:
a) \(2^7+2^9⋮10\)
Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)
\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)
\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)
\(\Leftrightarrow\overline{C8}+\overline{D2}\)
\(\Leftrightarrow\overline{E0}\)
Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)
b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)
Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)
\(\Leftrightarrow2^{72}.5^{20}\)
Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)
c) \(3^{10}+3^{12}⋮30\)
Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)
\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)
\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)
\(\Leftrightarrow\overline{C9}+\overline{B1}\)
\(\Leftrightarrow\overline{D0}⋮10\)
(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)
các bạn giải giúp mình với thứ 3 là mình phải trả bài rồi!
a/(-5/7)^n+1/(-5/7)^n với n >=1
b/(2x-3)^3=16
c/(3x-2)^5=-234
d/(0,8)^5/(0,4)^6
e/2^15.9^4/6^3.8^3
f/8^10+4^10/8^4+4^11
các bạn giúp gửi câu trả lời về nhanh cho mình nhé!^^
nếu bạn biết bài nào giúp tớ với nhé!!
Ta có : \(\left|2x+4\right|+\left|4x+8\right|=0\left|2x+4\right|+\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|+2.\left|2x+4\right|=\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|\left(1+2\right)=0\)
=> |2x + 4| = 0
=> 2x + 4 = 0
=> 2x = -4
=> x = -2
1. Đề đúng phải là thế này: \(\left|2x+4\right|+\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|=\left|4x+8\right|=0\)
\(\Rightarrow2x+4=4x+8=0\)
\(\Rightarrow x=-\frac{4}{2}=-\frac{8}{4}\)
\(\Rightarrow x=-2\)
2. Sửa lại đề : \(\left|x-5\right|-\left|x-7\right|=0\)
\(\Rightarrow\left|x-5\right|=\left|x-7\right|\)
\(\Rightarrow\orbr{\begin{cases}x-5=x-7\\x-5=-\left(x-7\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-5=-7\\x-5=-x+7\end{cases}}\)
( Loại trường hợp 1)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
3. \(\left|x+8\right|-\left|2x+2\right|=0\)
\(\Rightarrow\left|x+8\right|=\left|2x+2\right|\)
\(\Rightarrow\orbr{\begin{cases}x+8=2x+2\\x+8=-\left(2x+2\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+2=8\\x+8=-2x-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\3x=-10\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-\frac{10}{3}\end{cases}}\)