Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, Tự chép đề bài :v
=> 22x-3 = ( 83. 165 ) : 410
22x-3 = ( 29. 220 ) : 220
22x-3 = 229 : 220
22x-3 = 29
=> 2x - 3 = 9
2x = 9 + 3
2x = 12
x = 6
Vậy....
b, 7. 2x = 29 + 5. 28
7. 2x = 1792
2x = 1792 : 7
2x = 256
2x = 28
=> x = 8
Vậy ....
a \(\frac{2^{2x-3}}{4^{10}}=8^3.16^5\)
\(\Leftrightarrow\frac{2^{2x-3}}{4^{10}}=2^{29}\)
\(\Leftrightarrow2^{2x-3}=2^{29}.4^{10}\)
\(\Leftrightarrow2^{2x-3}=2^{49}\)
\(\Leftrightarrow2x-3=49\)
\(\Leftrightarrow x=26\)
b \(7.2^x=2^9+5.2^8\)
\(\Leftrightarrow7.2^x=2^8.(2+5)\)
\(\Leftrightarrow7.2^x=7.2^8\)
\(\Leftrightarrow x=7\)
Nếu phân số thứ 2 là \(\frac{1}{10.17}\) thì làm như vậy nè
\(\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{73.80}-\frac{1}{2.9}-\frac{1}{9.16}-\frac{1}{16.23}-\frac{1}{23.30}\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}\right)-\left(\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+\frac{1}{23.30}\right)\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}\right)-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)
= \(\frac{1}{7}.\frac{77}{240}-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{30}\right)=\frac{1}{7}.\frac{77}{240}-\frac{1}{7}.\frac{7}{15}\)
= \(\frac{11}{240}-\frac{1}{15}\)
= \(-\frac{1}{48}\)
a) Ta có: \(A=\left(\dfrac{63}{9\cdot18}+\dfrac{21}{14\cdot17}\right):\left(\dfrac{14}{9\cdot13}+\dfrac{14}{14\cdot18}+\dfrac{14}{3\cdot17}\right)\)
\(=\left(7\cdot\dfrac{9}{9\cdot18}+7\cdot\dfrac{3}{14\cdot17}\right):\left(14\left(\dfrac{1}{9\cdot13}+\dfrac{1}{14\cdot18}+\dfrac{1}{3\cdot17}\right)\right)\)
\(=\dfrac{7\left(\dfrac{1}{9}-\dfrac{1}{18}+\dfrac{1}{14}-\dfrac{1}{17}\right)}{14\cdot\dfrac{1}{4}\cdot\left(\dfrac{4}{9\cdot13}+\dfrac{4}{3\cdot17}+\dfrac{4}{14\cdot18}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{18}+\dfrac{1}{14}-\dfrac{1}{17}}{\dfrac{1}{2}\left(\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{3}-\dfrac{1}{17}+\dfrac{1}{14}-\dfrac{1}{18}\right)}\)
\(=\dfrac{\dfrac{73}{1071}}{\dfrac{1}{2}\cdot\dfrac{4519}{13923}}=\dfrac{1898}{4519}\)
\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)
\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)
\(\Rightarrow B=-\frac{113}{960}\)
\(C=0\)
\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
\(\Rightarrow D=1\)
D= \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}......-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
=\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{98}-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\left[1-(\frac{1}{2}-\frac{1}{2}+......+\frac{1}{98}-\frac{1}{99})\right]\)
=\(\frac{1}{99}-\left(1-0-0-.....-0-\frac{1}{99}\right)\)
=\(\frac{1}{99}-1-\frac{1}{99}\)
=1
a) Ta có:
\(\begin{array}{l}\frac{6}{{10}} = \frac{{6:2}}{{10:2}} = \frac{3}{5};\\\frac{9}{{15}} = \frac{{9:3}}{{15:3}} = \frac{3}{5}\end{array}\)
\(\begin{array}{l}\frac{{6 + 9}}{{10 + 15}} = \frac{{15}}{{25}} = \frac{{15:5}}{{25:5}} = \frac{3}{5};\\\frac{{6 - 9}}{{10 - 15}} = \frac{{ - 3}}{{ - 5}} = \frac{3}{5}\end{array}\)
Ta được: \(\frac{{6 + 9}}{{10 + 15}} = \frac{{6 - 9}}{{10 - 15}} = \frac{6}{{10}} = \frac{9}{{15}}\)
b) - Vì \(k = \frac{a}{b} \Rightarrow a = k.b\)
Vì \(k = \frac{c}{d} \Rightarrow c = k.d\)
- Ta có:
\(\begin{array}{l}\frac{{a + c}}{{b + d}} = \frac{{k.b + k.d}}{{b + d}} = \frac{{k.(b + d)}}{{b + d}} = k;\\\frac{{a - c}}{{b - d}} = \frac{{k.b - k.d}}{{b - d}} = \frac{{k.(b - d)}}{{b - d}} = k\end{array}\)
- Như vậy, \(\frac{{a + c}}{{b + d}}\) =\(\frac{{a - c}}{{b - d}}\) = \(\frac{a}{b}\) =\(\frac{c}{d}\)( = k)
a: \(\dfrac{6+9}{10+15}=\dfrac{15}{25}=\dfrac{3}{5};\dfrac{6-9}{10-15}=\dfrac{-3}{-5}=\dfrac{3}{5}\)
=>Bằng nhau
b: a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k;\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a}{b}=\dfrac{c}{d}\)
Bài làm
\(\frac{11}{15}-\frac{9}{10}< x< \frac{11}{15}:\frac{9}{10}\)
\(\Rightarrow\frac{22}{30}-\frac{27}{30}< x< \frac{11}{15}.\frac{10}{9}\)
\(\Rightarrow-\frac{5}{30}< x< \frac{11}{3}.\frac{2}{9}\)
\(\Rightarrow-\frac{5}{30}< x< \frac{22}{27}\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21\right\}\)
Vậy \(x\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21\right\}\)
~ Chắc z ~
# Chúc bạn học tốt #
Ta có:\(\frac{11}{15}-\frac{9}{10}< x< \frac{11}{15}:\frac{9}{10}\)
\(\Leftrightarrow\frac{110-135}{30}< x< \frac{11.10}{15.9}\)
\(\Leftrightarrow\frac{-15}{30}< x< \frac{22}{27}\)
(Vì x c Z)\(\Leftrightarrow-1< x< 1\Rightarrow x\in\left\{0\right\}\)
\(20\frac{15}{16}.16\frac{8}{9}-20\frac{15}{16}.12\frac{8}{9}\)
\(=20\frac{15}{16}.\left(16\frac{8}{9}-12\frac{8}{9}\right)\)
\(=20\frac{15}{16}.4\)
\(=83\frac{3}{4}\)
= 2/3 bn nhé
T mk nha!!!
\(\frac{3.17-9}{3.16+15}\)
=\(\frac{3.\left(16+1\right)-9}{3.16+15}\)
=\(\frac{3.16+3.1-9}{3.16+15}\)
=\(\frac{3.16+3-9}{3.16+15}\)
=\(\frac{3-9}{15}\)(vì đã triệt tiêu 3.16)
=\(\frac{-6}{15}\)=\(\frac{-2}{5}\)