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Ta có:
\(\frac{x+1}{98}+1+\frac{x+2}{97}+1=\frac{x+3}{96}+1+\frac{x+4}{95}+1\)
\(\frac{x+1}{98}+\frac{98}{98}+\frac{x+2}{97}+\frac{97}{97}=\frac{x+3}{96}+\frac{96}{96}+\frac{x+4}{95}+\frac{95}{95}\)
\(\frac{x+99}{98}+\frac{x+99}{97}=\frac{x+99}{96}+\frac{x+99}{95}\)
\(\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)
\(\left(x+99\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Vì: \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)nên x+99=0
=> x=-99
1/2 + 2/3 + 3/4 + 4/5 + 5/6 + 6/7 + 7/8 + 8/9 + ........+ 95/96 + 96/97 + 97/98 + 98/99 + 99/100 = ?
Số các số hạng là:
(2000 - 100) : 1 + 1 = 1901
Tổng là:
(2000 + 100) x 1901 : 2 = 1996050
Đáp số : 1996050
A = ( 4/4 + 2/3 ) - ( 51/3 - 6/5 ) - ( 6 - 7/4 + 3/2 )
Sau đó quy đồng rồi trừ cả là đc
B tương tự
C=13/15
D cx thế . Bạn tự vận dụng đi . Xl vì ko giải đc . Mik đang gấp
Các đề bài trên khi chuyển vế đều bị mất đi x nên không có x thỏa mãn
A=100+98+96+...+2−97−95−...1A=100+98+96+...+2−97−95−...1
A=100+(98−97)+(96−95)+...(2−1)A=100+(98−97)+(96−95)+...(2−1)
A=100+1+1+1+...+1A=100+1+1+1+...+1
A=100+1.49A=100+1.49
A=100+49A=100+49
A=149
a, 100 + 98 + 96 + ... + 2 - 9 7 - 95 - .. -1
= 100 + (98 - 97) + (96-95) + ... + + ... + (2 - 1)
= 100 + 1 + 1 + 1 +.. +1
= 100 + 1 x 49
= 100 + 49
= 149
b , 1 + 2 - 3 - 4 + 5 + 6 - .... -299 - 330 +301 + 302
=( 1 + 2 - 3) + ( -4 + 5 + 6 -7 ) +... +(298 - 299 -300 +301 ) + 302
= 0 + 0 + .. + 0 + 302
= 302
Bài 1:
a, \(A=3^{100}+3^{99}+...+3+1\)
\(\Rightarrow3A=3^{101}+3^{100}+...+3^2+3\)
\(\Rightarrow3A-A=\left(3^{101}+3^{100}+...+3^2+3\right)-\left(3^{100}+3^{99}+...+3+1\right)\)
\(\Rightarrow2A=3^{101}+1\Rightarrow A=\dfrac{3^{101}+1}{2}\)
b, \(B=\dfrac{15^9.2^{18}.9^8}{3^{15}.4^8.25^4}=\dfrac{3^9.5^9.2^{18}.3^{16}}{3^{15}.2^{16}.5^8}\)
\(=3^{10}.5.2^2=472392\)
c, \(C=\dfrac{2^{10}.10^{17}.7^9}{5^{15}.14^9.64^9}=\dfrac{2^{10}.2^{17}.5^{17}.7^9}{5^{15}.2^9.7^9.2^{54}}\)
\(=\dfrac{5^2}{2^{36}}\)
Chúc bạn học tốt!!!
1.
\(A=3^{100}+3^{99}+3^{98}+...+3^2+3+1\\ A=\dfrac{3-1}{2}\cdot\left(3^{100}+3^{99}+3^{98}+...+3^2+3+1\right)\\ =\dfrac{\left(3-1\right)\cdot\left(3^{100}+3^{99}+3^{98}+...+3^2+3+1\right)}{2}\\ =\dfrac{3^{101}-3^{100}+3^{100}-3^{99}+...+3^2-3+3-1}{2}\\ =\dfrac{3^{101}-1}{2}\)
\(B=\dfrac{15^9\cdot2^{18}\cdot9^8}{3^{15}\cdot4^8\cdot25^4}\\ =\dfrac{\left(3\cdot5\right)^9\cdot2^{18}\cdot\left(3^2\right)^8}{3^{15}\cdot\left(2^2\right)^8\cdot\left(5^2\right)^4}\\ =\dfrac{3^9\cdot5^9\cdot2^{18}\cdot3^{16}}{3^{15}\cdot2^{16}\cdot5^8}\\ =\dfrac{3^9\cdot5\cdot2^2\cdot3}{1\cdot1\cdot1}\\ =3^{10}\cdot5\cdot2^2\\ =59049\cdot5\cdot4\\ =59049\cdot\left(5\cdot4\right)\\ =59049\cdot20\\ =1180980\)
\(C=\dfrac{2^{10}\cdot10^{17}\cdot7^9}{5^{15}\cdot14^9\cdot64^9}\\ =\dfrac{2^{10}\cdot\left(2\cdot5\right)^{17}\cdot7^9}{5^{15}\cdot\left(2\cdot7\right)^9\cdot\left(2^6\right)^9}\\ =\dfrac{2^{10}\cdot2^{17}\cdot5^{17}\cdot7^9}{5^{15}\cdot2^9\cdot7^9\cdot2^{54}}\\ =\dfrac{2\cdot1\cdot5^2\cdot1}{1\cdot1\cdot1\cdot2^{37}}\\ =\dfrac{5^2}{2^{36}}\\ =\dfrac{25}{2^{36}}\)
1.a)\(2.x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2.x=\dfrac{20}{15}+\dfrac{5}{4}=\dfrac{4}{3}+\dfrac{5}{4}=\dfrac{16+15}{12}=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{12}:2=\dfrac{31}{12}.\dfrac{1}{2}=\dfrac{31}{24}\)
b)\(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{8}\right)\)
\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}=-\dfrac{5}{6}\)
2.Theo đề bài, ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\) và \(a+b=-15\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{-15}{5}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=-3\Rightarrow a=-6\\\dfrac{b}{3}=-3\Rightarrow b=-9\end{matrix}\right.\)
3.Ta xét từng trường hợp:
-TH1:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow x\in\left\{0;1\right\}\)
-TH2:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\left\{0;1\right\}\)
4.\(B=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^9=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^9=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{18}=\left(\dfrac{3}{7}\right)^3=\dfrac{27}{343}\)