VT
3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
18 tháng 6 2017
1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4
PN
3 tháng 7 2017
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...)
PN
3 tháng 7 2017
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)
DH
16 tháng 7 2017
a, \(x^2-9=0\Rightarrow x^2=9\Rightarrow x\pm3\)
b, \(\left(x-3\right)^2-25=0\Rightarrow\left(x-3\right)^2=25\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
c, \(\left(x-3\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
d, \(\left(x-3\right)x-2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
e, \(3x\left(x-1\right)-5\left(1-x\right)=0\)
\(\Rightarrow3x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
g, \(x^2+6x-7=0\)
\(\Rightarrow x^2-x+7x-7=0\)
\(\Rightarrow x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
h,\(2x^2+5x-7=0\)
\(\Rightarrow2x^2-2x+7x-7=0\)
\(\Rightarrow2x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Chúc bạn học tốt!!!
16 tháng 7 2017
a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) vậy \(x=3;x=-3\)
b) \(\left(x-3\right)^2-25=0\Leftrightarrow\left(x-3\right)^2=25\Leftrightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
vậy \(x=8;x=-2\)
c) \(\left(x-3\right)\left(2x-5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
vậy \(x=3;x=\dfrac{5}{2}\)
d)\(\left(x-3\right).x-2\left(x-3\right)=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) vậy \(x=2;x=3\)
e) \(3x\left(x-1\right)-5\left(1-x\right)=0\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-5}{3};x=1\)
câu e t thấy sai sai nhưng vẫn làm ; bn coi lại đề nha
g) \(x^2+6x-7=0\Leftrightarrow x^2-x+7x-7=0\)
\(\Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\) vậy \(x=-7;x=1\)
h) \(2x^2+5x-7=0\Leftrightarrow2x^2-2x+7x-7=0\)
\(\Leftrightarrow2x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(2x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-7}{2};x=1\)
10 tháng 2 2019
\(\frac{x-2}{4}=\frac{-9}{2-x}\)
\(\Rightarrow\frac{x-2}{4}=\frac{9}{x-2}\)
\(\Rightarrow\left(x-2\right)^2=36\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}}\)
\(\frac{x}{15}=\frac{3}{y}\)
\(\Rightarrow xy=45\)
\(\Rightarrow x;y\inƯ\left(45\right)=\left\{\pm1;\pm3;\pm5;\pm9;\pm15;\pm45\right\}\)
Xét bảng
| x | 1(loại) | -1 | 3(loại) | -3 | 5(loại) | -5 | 45 | -45(loại) | 15 | -15(loại) | 9 | -9(loại) |
| y | 45(loại) | -45 | 15(loại) | -15 | 9(loại) | -9 | 1 | -1(loại) | 3 | -3(loại) | 5 | -5(loại) |
Vậy.......................................
d;Áp dụng tích chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{4}=\frac{y}{3}=\frac{x+y}{4+3}=\frac{14}{7}=2\)
\(\Rightarrow x=4.2=8\)
\(y=3.2=6\)
A
20 tháng 8 2020
a, \(\frac{x}{5}=\frac{2}{3}\Leftrightarrow x=\frac{10}{3}\)
b, \(\frac{x}{-24}=\frac{20}{42}\Leftrightarrow x=-\frac{80}{7}\)
c, \(\frac{x+3}{15}=\frac{1}{3}\Leftrightarrow3x+9=15\Leftrightarrow x=2\)
d, \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\Leftrightarrow x\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\Leftrightarrow-\frac{5}{6}x=\frac{5}{12}\Leftrightarrow x=-\frac{1}{2}\)
27 tháng 9 2017
a, \(5^x=625\Rightarrow5^x=5^4\Rightarrow x=4\)
b, \(4^{2x-6}=1\Rightarrow4^{2x-6}=4^0\)
\(\Rightarrow2x-6=0\Rightarrow x=3\)
c, \(\left(3x-1\right)^3=8\Rightarrow3x-1=2\)
\(\Rightarrow x=1\)
d, \(49.7^n=2401\Rightarrow7^{n+2}=7^4\)
\(\Rightarrow n+2=4\Rightarrow n=2\)
e, \(x^4.x-27.x=0\)
\(\Rightarrow x\left(x^4-27\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^4-27=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt[4]{27}\end{cases}}\)
f, \(\left(x-6\right)^3=\left(x-6\right)^2\)
\(\Rightarrow\left(x-6\right)^3-\left(x-6\right)^2=0\)
\(\Rightarrow\left(x-6\right)^2\left(x-6-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
Chúc bạn học tốt!!!
M
27 tháng 9 2017
\(a,5^x=625\) \(b,4^{2x-6}=1\)
\(\Rightarrow5^x=5^4\) \(\Rightarrow2x-6=0\) (Vì mọi số mũ không bằng 1)
\(\Rightarrow x=4\) \(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
\(c,\left(3x-1\right)^3=8\) \(d,49.7^n=2401\)
\(\Rightarrow\left(3x-1\right)^3=2^3\) \(\Rightarrow7^n=2401:49\)
\(\Rightarrow3x-1=2\) \(\Rightarrow7^n=49\)
\(\Rightarrow3x=3\) \(\Rightarrow7^n=7^2\)
\(\Rightarrow x=1\) \(\Rightarrow n=2\)
\(e,x^4.x-27.x=0\)
\(\Rightarrow x.\left(x^4-27\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^4-27=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^4=27\end{cases}}}\)
\(f,\left(x-6\right)^3=\left(x-6\right)^2\)
\(\Rightarrow x.\left(x^4-27\right)=0\) \(\Rightarrow\hept{\begin{cases}\left(x-6\right)^3=\left(x-6\right)^2\\\left(x-6\right)^3=\left(x-6\right)^2\end{cases}\Rightarrow\hept{\begin{cases}\left(x-6\right)^3:\left(x-6\right)^2=1\\\left(x-6\right)^3-\left(x-6\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x-6=1\\x-6=0\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\x=6\end{cases}}}\)
KN
11 tháng 8 2020
a. Vì \(\left|x-y-5\right|\ge0\forall x;y;2019\left|y-3\right|^{2020}\ge0\forall y\)
\(\Rightarrow\left|x-y-5\right|+2019\left|y-3\right|^{2020}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-y-5\right|=0\\2019\left|y-3\right|^{2020}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y-5=0\\y-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=5\\y=3\end{cases}}\)
b. \(2\left(x-5\right)^4\ge0\forall x;5\left|2y-7\right|^5\ge0\forall y\)
\(\Rightarrow2\left(x-5\right)^4+5\left|2y-7\right|^5\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}2\left(x-5\right)^4=0\\5\left|2y-7\right|^5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-5=0\\2y-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\y=\frac{7}{2}\end{cases}}\)
d)x=0 hoặc x2-100=0
x2=100
x=10
e)x-1=0 hoặc x3+27=0
x=1 x3=27
x=3