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a . 254 x 399 - 145 / 254 + 399 x 253 = ( 253 + 1 ) x 399 - 145 / 254 + 399 x 253
= 253 x 399 + 1 x 399 - 145 / 254 + 399 x 253
=253 x 399 + 399 - 145 / 254 + 399 x 253
= 253 x 399 + ( 399 - 145) / 254 + 399 x 253
= 253 x 399 + 254 / 254 + 399 x 253
= 1
b. 1997 x 1996 - 995 / 1995 x 1997 + 1002 = 1997 x ( 1995 + 1 ) - 995 / 1995 x 1997 + 1002
= 1997 x 1995 + 1997 - 1995 / 1995 x 1997 + 1002
= 1997 x 1995 + ( 1997 - 995 ) / 1995 x1997 + 1002
= 1997 x1995 + 1002 / 1995 x 1997 + 1002
= 1
c. 5932 + 6001 x 5031 / 5932 x 6001 - 69 = 5932 + 6001 x 5031 / ( 5031 + 1 ) x 6001 - 69
= 5932 + 6001 x 5031 / 5031 x 6001 + 6001 - 69
= 5932 + 6001 x 5031 / 5031 x 6001 + ( 6001 - 69 )
= 5932 + 6001 x 5031 / 5031 x 6001 + 5932
= 1
d. 1995 x 1997 - 1 / 1996 x 1995 + 1994 = 1995 x ( 1996 + 1 ) - 1 / 1996 x 1995 + 1994
= 1995 x1996 + 1995 - 1 / 1996 x 1995 + 1994
= 1995 x 1996 + ( 1995 - 1 ) / 1996 x 1995 + 1994
= 1995 x 1996 + 1994 / 1996 x 1995 + 1994
= 1
\(\dfrac{29}{30}-\left(\dfrac{13}{30}+x\right)=\dfrac{7}{69}\)
\(\Leftrightarrow\dfrac{13}{23}+x=\dfrac{29}{30}-\dfrac{7}{69}\)
\(\Leftrightarrow\dfrac{13}{23}+x=\dfrac{199}{230}\)
\(\Leftrightarrow x=\dfrac{199}{230}-\dfrac{13}{23}\)
\(\Leftrightarrow x=\dfrac{69}{230}\)
Sửa đề:
Nếu:
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{69^{2015}+1}{69^{2017}+1}< 1\)
\(B< \dfrac{69^{2015}+1+68}{69^{2017}+1+68}\Leftrightarrow B< \dfrac{69^{2015}+69}{69^{2017}+69}\)
\(B< \dfrac{69\left(69^{2014}+1\right)}{69\left(69^{2016}+1\right)}\Leftrightarrow B< \dfrac{69^{2014}+1}{69^{2016}+1}=A\)
\(B< A\)
Ta có:
\(\dfrac{21}{3\cdot11}>\dfrac{12}{3\cdot11}\)
\(\dfrac{45}{11\cdot19}>\dfrac{12}{11\cdot19}\)
\(\dfrac{69}{19\cdot27}>\dfrac{12}{19\cdot27}\)
\(\Rightarrow\dfrac{21}{3\cdot11}-\dfrac{45}{11\cdot19}+\dfrac{69}{19\cdot27}>\dfrac{12}{3\cdot11}+\dfrac{12}{11\cdot19}+\dfrac{12}{19\cdot27}\)
\(\Rightarrow A>B\)
Vậy \(A>B\).
\(\dfrac{5932+6001.\left(5932-1\right)}{5932.6001-69}=\dfrac{5932+6001.5932-6001}{5932.6001-69}=\dfrac{6001.5932-69}{5932.6001-69}=1\)
\(\dfrac{5932+6001.5931}{5932.6001-69}\)
\(=\dfrac{5932+6001.5931}{\left(5931+1\right).6001-69}\)
\(=\dfrac{5932+6001.5931}{5931.6001+6001-69}\)
\(=\dfrac{5932}{6001-69}\)
\(=\dfrac{5932}{5932}\)
\(=1\)
P/s : Mình làm có chút vắn tắt. Xin bạn thông cảm
1.
\(\dfrac{19.20}{19+20}=\dfrac{380}{39}=9\dfrac{29}{39}\)
\(\dfrac{\overline{aaa}}{\overline{aa}}=\dfrac{111.a}{11.a}=\dfrac{111}{11}=10\dfrac{1}{11}\)
\(\dfrac{\overline{ababa}}{\overline{aba}}=\dfrac{100.\overline{aba}+\overline{ba}}{\overline{aba}}=\dfrac{100.\overline{aba}}{\overline{aba}}+\dfrac{\overline{ba}}{\overline{aba}}=100\dfrac{\overline{ba}}{\overline{aba}}\)
2.
\(6\dfrac{23}{41}=\dfrac{6.41+23}{41}=\dfrac{269}{41}\)
\(a\dfrac{a}{99}=\dfrac{a.99+a}{99}=\dfrac{100.a}{99}=\dfrac{\overline{a00}}{99}\)
\(1\dfrac{a-b}{a+b}=\dfrac{a+b+a-b}{a+b}=\dfrac{2.a}{a+b}\)
3.
\(\dfrac{69}{1000}=0,069\)
\(8\dfrac{77}{100}=8,77\)
\(\dfrac{34567}{10^4}=\dfrac{34567}{10000}=3,4567\)
\(\dfrac{\overline{abc}}{10^n}=\dfrac{\overline{abc}}{10...0}=\overline{0,0...0abc}\)
n số hạng 0 n - 3 số hạng 0 ở phần thập phân
a ) \(7x=4y\) hay \(\dfrac{x}{4}=\dfrac{y}{7}\) và \(y-z=24\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{y-z}{7-4}=\dfrac{24}{3}=8\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=32\\y=56\end{matrix}\right.\)
Vậy ............
b ) \(\dfrac{x}{5}=\dfrac{y}{6},\dfrac{y}{8}=\dfrac{z}{7}\)
hay : \(\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}\) và \(x+y-z=69\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)
Vậy .......
c.
\(\dfrac{x}{y}=\dfrac{2}{5}=\dfrac{x}{2}=\dfrac{y}{5}\)và x - y = 40
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x-y}{2-5}=\dfrac{40}{-3}\)
\(\dfrac{x}{2}=\dfrac{40}{-3}\Rightarrow x=\dfrac{40.2}{-3}=-\dfrac{80}{3}\)
\(\dfrac{y}{5}=\dfrac{40}{-3}\Rightarrow y=\dfrac{40.5}{-3}=-\dfrac{200}{3}\)
Vậy x = \(-\dfrac{80}{3}\), y = \(-\dfrac{200}{3}\)
Tương tự tiếp nghen
\(N=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot5\cdot2^{18}}{3^9\cdot2^9\cdot2^{10}-2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\cdot\left(2-5\right)}{3^9\cdot2^{19}\left(1-2\cdot3\right)}=\dfrac{1}{2}\cdot\dfrac{-3}{-5}=\dfrac{3}{10}\)
\(P=\dfrac{8\cdot10+8\cdot24+8\cdot560}{6\cdot45+6\cdot108+6\cdot120\cdot21}=\dfrac{8\left(10+24+560\right)}{6\left(45+108+120\cdot21\right)}=\dfrac{4}{3}\cdot\dfrac{2}{9}=\dfrac{8}{27}\)