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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) \(\left(19x+2\times5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Rightarrow\left(19x+50\right):14=5^2-4^2=25-16=9\)
\(\Rightarrow19x+50=126\)
\(\Rightarrow19x=76\Rightarrow x=4\)
Vậy x = 4
b) \(2\times3^2=10\times3^{12}+8\times27^4\)
\(\Rightarrow2\times3^2=10\times\left(3^3\right)^4+8\times27^4\)
\(\Rightarrow2\times3^2=27^4\times\left(10+8\right)\)
\(\Rightarrow18=27^4\times18\)
\(\Rightarrow27^4\times18-18=0\Rightarrow18\times\left(27^4-1\right)=0\)
=> Không thấy biến x đâu cả
c) Ta thấy 33 = 27
\(\Rightarrow3^{2x-5}=3^3\Rightarrow2x-5=3\Rightarrow2x=8\Rightarrow x=4\)
Vậy x = 4
d) \(3^{x+1}-x=80\Rightarrow3^{x+1}=81\)
Ta thấy 34 = 81
\(\Rightarrow3^{x+1}=3^4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
a) (5x - 1) : 3 + 1 = 4
=> (5x - 1) : 3 = 3
=> (5x - 1) = 9
=> 5x - 1 = 9
=> 5x = 10
=> x = 2
b) 54 : (16 - x) - 1=5
=> 54:(16-x) = 6
=> 16-x = 9
=> x = 7
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)
a)437+(248-x)=635
248-x=635-437=198
x=248-198=50
b)5.(x+27)=465
x+27=465/5=93
x=93-27=66
c)3.(x+3)=30
x+3=30:3=10
x=10-3=7
d)8x-9=31
8x=31-9=22
x=22/8=11/4
e)7x-9=26
7x=26+9=35
x=35:7=5
g)76-3.(x+5)=16
3.(x+5)=76-16=60
x+5=60/3=20
x=15
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
Câu 1:
\(A=\frac{\left(1+2+3+...+100\right)x\left(101x102-101x101-51-50\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x\left(101x\left(102-101\right)-\left(50+51\right)\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x\left(101-101\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x0}{2+4+6+8+...+2048}\)
\(A=0\)
Ta có:Số số hạng từ 2 đến 101 là:
(101-2):1+1=100(số hạng)
Do đó từ 2 đến 101 có số cặp là:
100:2=50(cặp)
\(B=\frac{101+100+99+...+3+2+1}{101-100+99-98+3-2+1}\)
\(B=\frac{5151}{51}\)
\(B=101\)
Câu 2:
a)697:\(\frac{15x+364}{x}\)=17
\(\frac{15x+364}{x}\)=697:17
\(\frac{15x+364}{x}\)=41
15x+364=41x
41x-15x=364
26x=364
x=14
Vậy x=14
b)92.4-27=\(\frac{x+350}{x}+315\)
\(\frac{x+350}{x}+315\)=341
\(\frac{x+350}{x}\)=26
x+350=26
x=26-350
x=-324
Vậy x=-324
c, 720 : [ 41 - ( 2x -5)] = 40
[ 41 - ( 2x -5)] =720:40
[ 41 - ( 2x -5)] =18
2x-5=41-18
2x-5=23
2x=28
x=14
Vậy x=14
d, Số số hạng từ 1 đến 100 là:
(100-1):1+1=100(số hạng)
Tổng dãy số là:
(100+1)x100:2=5050
Mà cứ 1 số hạng lại có 1x suy ra có 100x
Ta có:(x+1) + (x+2) +...+ (x+100) = 5750
(x+x+...+x)+(1+2+...+100)=5750
100x+5050=5750
100x=700
x=7
Vậy x=7
a) \(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
b) \(2^x.16=128\)
\(2^x=128:16\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
c) \(3^x:9=27\)
\(3^x=27.9\)
\(3^x=243\)
\(3^x=3^5\)
\(\Rightarrow x=5\)
d) \(x^4=x\)
\(\Rightarrow x=0\)hoac \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
e) \(\left(2x+1\right)^3=27\)
\(\left(2x+1\right)^3=3^3\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
f) \(\left(x-2\right)^2=\left(x-2\right)^4\)
\(\left(x-2\right)^2-\left(x-2\right)^4=0\)
\(\left(x-2\right)^2-\left(x-2\right)^2.\left(x-2\right)^2=0\)
\(\left(x-2\right)^2\left[1-\left(x-2\right)^2\right]=0\)
\(\left(x-2\right)^2\left(1-x+2\right)\left(1+x-2\right)=0\)
\(\Rightarrow\left(x-2\right)^2=0\)hoac \(\orbr{\begin{cases}3-x=0\\x-1=0\end{cases}}\)
\(\Rightarrow x-2=0\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
a) \(3^x=81\Leftrightarrow3^x=3^4\Rightarrow x=4\)
b)\(2^x\times16=128\Leftrightarrow2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
c) \(3^x\div9=27\Leftrightarrow3^x\div3^2=3^3\Rightarrow x=5\)
d) \(x^4=x\Leftrightarrow x=1\)
e) \(\left(2x+1\right)^3=27\Leftrightarrow\left(2x+1\right)^3=3^3\Rightarrow2x+1=3 \)
\(\Rightarrow2x=3+1\Leftrightarrow2x=4\Rightarrow x=2\)
F)
a.
\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{2}\)
\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{2}\)
\(-\frac{5}{6}\times x=\frac{5}{2}\)
\(x=\frac{5}{2}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{5}{2}\times\left(-\frac{6}{5}\right)\)
\(x=-3\)
b.
\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=\frac{-53-4}{10}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)
\(3x-3,7=-\frac{57}{10}\div\frac{3}{5}\)
\(3x-3,7=-\frac{57}{10}\times\frac{5}{3}\)
\(3x-\frac{37}{10}=-\frac{19}{2}\)
\(3x=-\frac{19}{2}+\frac{37}{10}\)
\(3x=\frac{-95+37}{10}\)
\(3x=-\frac{58}{10}\)
\(3x=-\frac{29}{5}\)
\(x=-\frac{29}{5}\div3\)
\(x=-\frac{29}{5}\times\frac{1}{3}\)
\(x=-\frac{29}{15}\)
c.
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23-15}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\times\frac{27}{8}\)
\(2+\frac{3}{4}x=\frac{21}{8}\)
\(\frac{3}{4}x=\frac{21}{8}-2\)
\(\frac{3}{4}x=\frac{21-16}{8}\)
\(\frac{3}{4}x=\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{3}{4}\)
\(x=\frac{5}{8}\times\frac{4}{3}\)
\(x=\frac{5}{6}\)
d.
\(-\frac{2}{3}\times x+\frac{1}{5}=\frac{3}{10}\)
\(-\frac{2}{3}\times x=\frac{3}{10}-\frac{1}{5}\)
\(-\frac{2}{3}\times x=\frac{3-2}{10}\)
\(-\frac{2}{3}\times x=\frac{1}{10}\)
\(x=\frac{1}{10}\div\left(-\frac{2}{3}\right)\)
\(x=\frac{1}{10}\times\left(-\frac{3}{2}\right)\)
\(x=-\frac{3}{20}\)
e.
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x\right|=\frac{20+9}{12}\)
\(\left|x\right|=\frac{29}{12}\)
\(x=\pm\frac{29}{12}\)
Vậy \(x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)
f.
\(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)
\(\left|2x-\frac{1}{3}\right|=1-\frac{5}{6}\)
\(\left|2x-\frac{1}{3}\right|=\frac{6-5}{6}\)
\(\left|2x-\frac{1}{3}\right|=\frac{1}{6}\)
\(2x-\frac{1}{3}=\pm\frac{1}{6}\)
- \(2x-\frac{1}{3}=\frac{1}{6}\)
\(2x=\frac{1}{6}+\frac{1}{3}\)
\(2x=\frac{1+2}{6}\)
\(2x=\frac{3}{6}\)
\(2x=\frac{1}{2}\)
\(x=\frac{1}{2}\div2\)
\(x=\frac{1}{2}\times\frac{1}{2}\)
\(x=\frac{1}{4}\)
- \(2x-\frac{1}{3}=-\frac{1}{6}\)
\(2x=-\frac{1}{6}+\frac{1}{3}\)
\(2x=\frac{-1+2}{6}\)
\(2x=\frac{1}{6}\)
\(x=\frac{1}{6}\div2\)
\(x=\frac{1}{6}\times\frac{1}{2}\)
\(x=\frac{1}{12}\)
Vậy x = 1/4 hoặc x = 1/12.
Chúc bạn học tốt
Sorry nha, mik chép lộn đề
Làm lại câu a nha
a.
\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{12}\)
\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{12}\)
\(-\frac{5}{6}\times x=\frac{5}{12}\)
\(x=\frac{5}{12}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{5}{12}\times\left(-\frac{6}{5}\right)\)
\(x=-\frac{1}{2}\)
Chúc bạn học tốt
a) x = 4
b) x = 3
c) x = 2
d) x = 1
e) x = 3
f) x = 2
g) x = 4
h) x = 3