Bài giải

1 chục = 10 . Vậy số chim bay đi là 10 con.

Số con chim trên cành còn :

             20 - 10 = 10 ( con )

                   Đáp số : 10 con

                                                                                    Giải

                                                                         Đổi : 1 chục = 10 

Còn số con chim trên cành là

    20 - 10 = 10 ( con chim )

                 Đáp số : 10 con chim

Doi :1 chuc=10 ;con lai so chim la: 20 - 10 = 10 (con)

dap so: 10 con chim

tk cho minh roi ket ban nha

Câu I:

1) Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{x+\sqrt{x}-4}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{x+\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\dfrac{x+x+\sqrt{x}-2-x-\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)

\(=\dfrac{x+2}{\sqrt{x}}\)

2) Để P=3 thì \(\dfrac{x+2}{\sqrt{x}}=3\)

\(\Leftrightarrow x+2=3\sqrt{x}\)

\(\Leftrightarrow x-3\sqrt{x}+2=0\)

\(\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\)

\(\Leftrightarrow\sqrt{x}\cdot\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy: Để P=3 thì x=4

1, set about: attack someone

2, call back: return to see or collect sth

3, pay me back: return money to someone you borrow

4, go right back to: kiểu nó bắt đầu từ thời Trung Cổ ấy

5, look after: chắc cái này bạn biết nhỉ

6, set about: start doing something

7, turn back: return

8, went about: begin to do something or deal with something

9, called me back: telephone someone again

10, holding back: stating sth

Bài 3:

Chiều rộng là:

\(\dfrac{11}{4}-\dfrac{3}{2}=\dfrac{11}{4}-\dfrac{6}{4}=\dfrac{5}{4}\left(m\right)\)

Chu vi là:

\(\left(\dfrac{11}{4}+\dfrac{5}{4}\right)\cdot2=\dfrac{16}{4}\cdot2=4\cdot2=8\left(m\right)\)

Diện tích là:

\(\dfrac{11}{4}\cdot\dfrac{5}{4}=\dfrac{55}{16}\left(m^2\right)\)

bài 1 :

a) \(3\dfrac{4}{5}\)= \(\dfrac{19}{5}\)                                   b) \(4\dfrac{3}{5}\)=\(\dfrac{23}{5}\)

c) \(11\dfrac{12}{13}\)= \(\dfrac{155}{13}\)                              d) 12 \(\dfrac{11}{13}\)= \(\dfrac{167}{13}\)

bài 1
a)
\(3\dfrac{4}{5}=\dfrac{3x5+4}{5}=\dfrac{19}{5}\)
b)
\(4\dfrac{3}{5}=\dfrac{4x5+3}{5}=\dfrac{23}{5}\)
c)
\(11\dfrac{12}{13}=\dfrac{11x13+12}{13}=\dfrac{155}{13}\)
d)
\(12\dfrac{11}{13}=\dfrac{12x13+11}{13}=\dfrac{167}{13}\)
bài 2
a)\(3\dfrac{3}{4}+1\dfrac{1}{2}=\dfrac{3x4+3}{4}+\dfrac{1x2+1}{2}=\dfrac{15}{4}+\dfrac{3}{2}=\dfrac{21}{4}\)
b)\(3\dfrac{3}{4}-1\dfrac{1}{2}=\dfrac{15}{4}-\dfrac{3}{2}=\dfrac{15}{4}-\dfrac{6}{4}=\dfrac{9}{4}\)
c)\(3\dfrac{3}{4}x1\dfrac{1}{2}=\dfrac{15}{4}x\dfrac{3}{2}=\dfrac{45}{8}\)
d)\(3\dfrac{3}{4}:1\dfrac{1}{2}=\dfrac{15}{4}:\dfrac{3}{2}=\dfrac{15}{4}.\dfrac{2}{3}=\dfrac{5}{2}\)

HC=16cm

AB=13cm

=>BC=21cm

=>C=13+20+21=54(cm)

viết câu trả lời dài dài đầy đủ giúp em:')

Xét \(\Delta AHC\) vuông tại H có:

     \(AC^2=AH^2+HC^2\) ( định lý Pytago)

      \(20^2=12^2+HC^2\)

      \(\Rightarrow HC^2=400-144\)

       \(HC^2=256\)

       HC=16 cm

Ta có \(HC+BH=BC\)

=>    \(16+5=21\) cm

Xét \(\Delta AHB\) vuông tại H có:

     \(AB^2=AH^2+HB^2\) ( định lý Pytago)

      \(AB^2=12^2+5^2\)

      \(\Rightarrow AB^2=144+25\)

       \(AB^2=169\)

=> AB= 13

Chu vi hình tam giác ABC là : AB + BC + AC = 13+21+20= 54 CM

Giải:

\(A=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\) 

\(A=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\) 

\(A=\dfrac{1}{6}.\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+\dfrac{6}{19.25}+\dfrac{6}{25.31}+\dfrac{6}{31.37}\right)\) 

\(A=\dfrac{1}{6}.\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\) 

\(A=\dfrac{1}{6}.\left(1-\dfrac{1}{37}\right)\) 

\(A=\dfrac{1}{6}.\dfrac{36}{37}\) 

\(A=\dfrac{6}{37}\)

\(6A=\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{31.37}\)

= \(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{31}-\dfrac{1}{37}=\dfrac{36}{37}\)

<=> A = \(\dfrac{6}{37}\)

Ta có: \(A=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)

\(=\dfrac{1}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+\dfrac{6}{19\cdot25}+\dfrac{6}{25\cdot31}+\dfrac{6}{31\cdot37}\right)\)

\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\cdot\dfrac{36}{37}=\dfrac{6}{37}\)

a) vì các đèn mắc song song nên

\(U=U_1=U_2=2,8V\)

b)vì các đèn mắc song song nên

\(I=I_1+I_2\)

\(=>I_1=I-I_2=0,45-0,22=0,23A\)