Đang cần gấp ạ
- 3.2+(-5.2)+1
- -4.9-1.3
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\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{-49}{50}\)
a) \(2^n=8\)
\(\Rightarrow2^n=2^3\)
\(\Rightarrow n=3\)
b) \(5^{n+1}=125\)
\(\Rightarrow5^{n+1}=5^3\)
\(\Rightarrow n+1=3\)
\(\Rightarrow n=3-1=2\)
c) Mình không rõ đề:
d) \(2\cdot7^{n-1}+3=101\)
\(\Rightarrow2\cdot7^{n-1}=101-3\)
\(\Rightarrow2\cdot7^{n-1}=98\)
\(\Rightarrow7^{n-1}=\dfrac{98}{2}\)
\(\Rightarrow7^{n-1}=49\)
\(\Rightarrow7^{n-1}=7^2\)
\(\Rightarrow n-1=2\)
\(\Rightarrow n=1+2=3\)
e) \(3\cdot5^{2n+1}-6^2=339\)
\(\Rightarrow3\cdot5^{2n+1}=339+36\)
\(\Rightarrow3\cdot5^{2n+1}=375\)
\(\Rightarrow5^{2n+1}=125\)
\(\Rightarrow5^{2n+1}=5^3\)
\(\Rightarrow2n+1=3\)
\(\Rightarrow2n=2\)
\(\Rightarrow n=\dfrac{2}{2}=1\)
1.2 với \(x\ge0,x\in Z\)
A=\(\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\in Z< =>\sqrt{x}+2\inƯ\left(3\right)=\left(\pm1;\pm3\right)\)
*\(\sqrt{x}+2=1=>\sqrt{x}=-1\)(vô lí)
*\(\sqrt{x}+2=-1=>\sqrt{x}=-3\)(vô lí
*\(\sqrt{x}+2=3=>x=1\)(TM)
*\(\sqrt{x}+2=-3=\sqrt{x}=-5\)(vô lí)
vậy x=1 thì A\(\in Z\)
a) Sửa đề: Chứng minh \(\Delta DAB\) \(\sim\) \(\Delta CBD\)
Xét \(\Delta DAB\) và \(\Delta CBD\) có:
\(\widehat{DAB}=\widehat{CBD}\left(gt\right)\)
\(\widehat{ABD}=\widehat{BDC}\) (so le trong)
\(\Rightarrow\Delta DAB\sim\Delta CBD\left(g-g\right)\)
b) Do \(\Delta DAB\sim\Delta CBD\) (cmt)
\(\Rightarrow\dfrac{AD}{BC}=\dfrac{AB}{BD}=\dfrac{BD}{CD}\)
*) \(\dfrac{AD}{BC}=\dfrac{AB}{BD}\)
\(\Rightarrow BC=\dfrac{AD.BD}{AB}\)
\(=\dfrac{4.6}{3}=8\left(cm\right)\)
*) \(\dfrac{AB}{BD}=\dfrac{BD}{CD}\)
\(\Rightarrow CD=\dfrac{BD.BD}{AB}\)
\(=\dfrac{6.6}{3}=12\left(cm\right)\)
giup mink ai nhanh tick 3 lan nghia la 1 lan tra loi mot lan noi lam rui mot lan noi cam on
1, 3.2 + ( -5.2 ) + 1
= 6 + ( - 10 ) + 1 = -3
2, -4.9 -1.3
= -36 - 3 =-39