Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)
9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
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!!!
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 \(x=\dfrac{-7}{2};x=1\)
6. \(\dfrac{x}{4}=\dfrac{9}{x}\)
=>x2=4.9=36
=>x\(\in\)\(\left\{-6;6\right\}\)
\((\dfrac{2x}{5}+2):\left(-4\right)=-1\dfrac{1}{2}\)
(\(\dfrac{2x}{5}+2):\left(-4\right)=-\dfrac{3}{2}\)
\(\dfrac{2x}{5}=-\dfrac{3}{2}.\left(-4\right)\)
\(\dfrac{2x}{5}=6\)
\(\dfrac{2x}{5}=\dfrac{30}{5}\)
2x = 30
x = 30 : 2 = 15
Ta có : \(\left(5x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\2x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=3\\2x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{5}{2}\end{cases}}\)
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{2}{3}x+\dfrac{2}{3}x\)
\(\Rightarrow\dfrac{1}{2}=\dfrac{4}{3}x\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\).
1) Ta có: \(3\left(x-1\right)-5\left(x-2\right)=4\left(x+1\right)\)
\(\Leftrightarrow3x-5-5x+10-4x-4=0\)
\(\Leftrightarrow-6x+1=0\)
\(\Leftrightarrow-6x=-1\)
hay \(x=\dfrac{1}{6}\)
2) Ta có: \(-2\left(x-2\right)-4\left(x+1\right)=-3\left(x+3\right)\)
\(\Leftrightarrow-2x+4-4x-4+3x+9=0\)
\(\Leftrightarrow-3x=-9\)
hay x=3
3) Ta có: \(3x^2+2x=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
4) Ta có: \(x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) Ta có: \(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
6) Ta có: \(\left(5x-1\right)^3=125\)
\(\Leftrightarrow5x-1=5\)
\(\Leftrightarrow5x=6\)
hay \(x=\dfrac{6}{5}\)
7) Ta có: \(3^{x+1}=27\)
\(\Leftrightarrow x+1=3\)
hay x=2