Đổi: 8 phút = \(\dfrac{2}{15}\left(h\right)\)

Vận tốc của xe đạp là:

\(1,6:\dfrac{2}{15}=12\left(km/h\right)\)

1,6km=1600m

Vận tốc của xe đạp là 1600:8=200(m/p)=12km/h

=>Chọn C

1: Sửa đề: Vẽ \(\widehat{x'Ay'}\) là góc đối đỉnh của góc xAy

2: Ta có: \(\widehat{xAy}+\widehat{xAy'}=180^0\)(hai góc kề bù)

=>\(\widehat{xAy'}+100^0=180^0\)

=>\(\widehat{xAy'}=80^0\)

Ta có: \(\widehat{xAy}=\widehat{x'Ay'}\)(hai góc đối đỉnh)

mà \(\widehat{xAy}=100^0\)

nên \(\widehat{x'A'y}=100^0\)

Ta có: \(\widehat{xAy'}=\widehat{x'Ay}\)(hai góc đối đỉnh)

mà \(\widehat{xAy'}=80^0\)

nên \(\widehat{x'Ay}=80^0\)

\(\left[\left(\dfrac{4}{3}\right)^{-3}\cdot\left(\dfrac{3}{4}\right)^5\right]:\left(\dfrac{3}{8}\right)^7\\ =\left[\left(\dfrac{3}{4}\right)^3\cdot\left(\dfrac{3}{4}\right)^5\right]:\left(\dfrac{3}{8}\right)^7\\ =\left(\dfrac{3}{4}\right)^{3+5}:\dfrac{3^7}{8^7}\\ =\left(\dfrac{3}{4}\right)^8\cdot\dfrac{8^7}{3^7}\\ =\dfrac{3^8}{4^8}\cdot\dfrac{8^7}{3^7}\\ =\dfrac{3^8}{2^{16}}\cdot\dfrac{2^{21}}{3^7}=3\cdot2^5=3\cdot32=96\)

\(\left[\left(\dfrac{4}{3}\right)^{-3}\cdot\left(\dfrac{3}{4}\right)^5\right]:\left(\dfrac{3}{8}\right)^7\)

\(=\left[\left(\dfrac{3}{4}\right)^3\cdot\left(\dfrac{3}{4}\right)^5\right]:\dfrac{3^7}{8^7}\)

\(=\left(\dfrac{3}{4}\right)^8\cdot\dfrac{8^7}{3^7}=\dfrac{3^8}{4^8}\cdot\dfrac{8^7}{3^7}=\dfrac{3\cdot2^{21}}{2^{16}}=3\cdot2^5=3\cdot32=96\)

Ai giúp em với ạ em đang cần gấp lắm

Hiệu của hai số khi tăng số lớn thêm 3 đơn vị và giảm số bé đi 2 đơn vị là:

35+3-(-2)=38+2=40

Số lớn khi đó là 40:4x5=50

Số lớn là 50-3=47

Số bé là 47-35=12

a: \(\left\{{}\begin{matrix}x+3y=11\\3x-y=9-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+3y=11\\3x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=11\\9x+3y=27\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x+3y-x-3y=27-11\\x+3y=11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}8x=16\\3y=11-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{11-x}{3}=\dfrac{11-2}{3}=\dfrac{9}{3}=3\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+10y-3x=-1\\2x+4-3x+15y=-12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+10y=-1\\-x+15y=-16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-2x+30y=-32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+10y-2x+30y=-1+\left(-32\right)\\x-15y=16\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}40y=-33\\x=15y+16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x=15\cdot\dfrac{-33}{40}+16=\dfrac{29}{8}\end{matrix}\right.\)

a) 

\(\left\{{}\begin{matrix}x+3y=11\\3x-y=9-2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+3y=11\\3x+y=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+9y=33\\3x+y=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}8y=24\\3x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\3x+3=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{6}{3}=2\end{matrix}\right.\)

b) 

\(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-15y-12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\x-15y=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\2x-30y=32\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}40y=-33\\x-15y=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x+\dfrac{99}{8}=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{33}{40}\\x=16-\dfrac{99}{8}=\dfrac{29}{8}\end{matrix}\right.\)

Đây không phải là toán lớp 4. Hình như bạn nhầm rồi

B={x\(\in\)N|x lẻ; 4<x<14}

=>B={5;7;9;11;13}

Bài 4.Trong các phép so sánh sau, phép so sánh nào sai?

A. 71,56 < 71,65   B. 71,506 > 71,056

C. 716,05 > 716,005   D. 716,50 < 716,500

Vì 716,50 = 716,5 ; 716, 500 = 716,5 nên 716,50 < 716,500 là sai.

Chọn D

Chu vi hình vuông ABCD là 56cm

=>4xAB=56cm

=>AB=56/4=14(cm)

\(AC=\sqrt{14^2+14^2}=14\sqrt{2}\left(cm\right)\)

=>\(OA=OC=\dfrac{14\sqrt{2}}{2}=7\sqrt{2}\left(cm\right)\)

Diện tích hình tròn tâm O là:

\(S_{\left(O\right)}=OA\times OA\times3,14=307,72\left(cm^2\right)\)

Diện tích hình vuông ABCD là:

\(S_{ABCD}=14\times14=196\left(cm^2\right)\)

Diện tích phần gạch chéo là:

307,72-196=111,72(cm²)

\(\dfrac{x+100}{4}+\dfrac{x+99}{5}=\dfrac{x+98}{6}+\dfrac{x+97}{7}\)

=>\(\left(\dfrac{x+100}{4}+1\right)+\left(\dfrac{x+99}{5}+1\right)=\left(\dfrac{x+98}{6}+1\right)+\left(\dfrac{x+97}{7}+1\right)\)

=>\(\dfrac{x+104}{4}+\dfrac{x+104}{5}=\dfrac{x+104}{6}+\dfrac{x+104}{7}\)

=>\(\left(x+104\right)\left(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)=0\)

=>x+104=0

=>x=-104

\(\dfrac{x+100}{4}+\dfrac{x+99}{5}=\dfrac{x+98}{6}+\dfrac{x+97}{7}\\ \dfrac{x+100}{4}+\dfrac{x+99}{5}-\dfrac{x+98}{6}-\dfrac{x+97}{7}=0\\ \left(\dfrac{x+100}{4}+1\right)+\left(\dfrac{x+99}{5}+1\right)-\left(\dfrac{x+98}{6}+1\right)-\left(\dfrac{x+97}{7}+1\right)=0\\ \dfrac{x+104}{4}+\dfrac{x+104}{5}-\dfrac{x+104}{6}-\dfrac{x+104}{7}=0\\ \left(x+104\right)\cdot\left(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)=0\)

Vì \(\left(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)\ne0\) nên:

\(x+104=0\\ x=-104\)

Vậy \(x=-104\)