Ai giúp em với ạ em cần rất gấp bài 4 thôi ạ
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Bài 5:
a: Xét ΔBEC và ΔADC có
\(\widehat{C}\) chung
\(\widehat{EBC}=\widehat{DAC}\)
Do đó: ΔBEC\(\sim\)ΔADC
1 Having slept
2 not being invited
3 Having had
4 having
5 talking
6 succeeded - launching
7 Having travelled
8 Have - considered - trying
9 Having seen - had - to go
10 Being invited
11 Being found
12 having
13 taken - being photographed
14 to fix
15 living
16 Having waited - to deliver - decided to cancel
17 Having photocopied
18 to have happen
19 to give
20 spoiling
3.15:
EF vuông góc MH
NP vuông góc MH
Do đó: EF//NP
3.17:
góc yKH+góc H=180 độ
mà hai góc này là hai góc ở vị trí trong cùng phía
nên Ky//Hx
what is her mother going to prepare for her bỉthdat party
bài 3:
tổng số giờ đã chảy đc từ 2 vòi : 1+1=2(giờ)
tổng số phần bể đã chảy được từ 2 vòi : \(\dfrac{1}{5}+\dfrac{1}{7}=\dfrac{7}{35}+\dfrac{5}{35}=\dfrac{12}{35}\left(ph\text{ần} b\text{ể}\right)\)
nếu chảy cùng lúc mỗi giờ chảy được : \(\dfrac{12}{35}:2=\dfrac{12}{35\cdot2}=\dfrac{6}{35}\left(ph\text{ần}b\text{ể}\right)\)
bài 4:
cách 1:
độ dài đoạn AB là : \(\dfrac{3}{4}+\dfrac{9}{8}=\dfrac{18}{24}+\dfrac{27}{24}=\dfrac{45}{24}\left(m\right)\)
diện tích ABCD là : \(\dfrac{45}{27}\cdot\dfrac{4}{7}=\dfrac{15}{14}\left(m^2\right)\)
cách 2:
diện tích AEFD là : \(\dfrac{3}{4}\cdot\dfrac{4}{7}=\dfrac{3}{7}\left(m^2\right)\)
diện tích EBCF là : \(\dfrac{9}{8}\cdot\dfrac{4}{7}=\dfrac{9}{14}\left(m^2\right)\)
diện tích ABCD là : \(\dfrac{3}{7}+\dfrac{9}{14}=\dfrac{15}{14}\left(m^2\right)\)
Bài 4:
a: \(\sqrt{1,6}\cdot\sqrt{250}+\sqrt{19,6}:\sqrt{4,9}\)
\(=\sqrt{1,6\cdot250}+\sqrt4\)
\(=\sqrt{400}+2=20+2=22\)
b: \(\sqrt{1\frac34\cdot2\frac27\cdot5\frac49}\)
\(=\sqrt{\frac74\cdot\frac{16}{7}\cdot\frac{49}{9}}=\sqrt{\frac{16}{4}\cdot\frac{49}{9}}=2\cdot\frac73=\frac{14}{3}\)
c: \(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)
\(=20\sqrt{20}-15\sqrt{45}+5\sqrt5\)
\(=40\sqrt5-45\sqrt5+5\sqrt5=0\)
d: \(\left(\sqrt{325}-\sqrt{117}+2\sqrt{208}\right):\sqrt{13}\)
\(=\sqrt{25}-\sqrt9+2\cdot\sqrt{16}\)
\(=5-3+2\cdot4\)
=2+8
=10
e: \(\frac{2\sqrt8-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt5+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\frac{4\sqrt2-2\sqrt3}{\sqrt6\left(\sqrt3-2\sqrt2\right)}-\frac{\sqrt5+\sqrt{27}}{\sqrt6\left(\sqrt5+\sqrt{27}\right)}\)
\(=\frac{2\left(2\sqrt2-\sqrt3\right)}{-\sqrt6\left(2\sqrt2-\sqrt3\right)}-\frac{1}{\sqrt6}=-\frac{2}{\sqrt6}-\frac{1}{\sqrt6}=-\frac{3}{\sqrt6}=\frac{-3\sqrt6}{6}=-\frac{\sqrt6}{2}\)
f: \(\frac{3+2\sqrt3}{\sqrt3}+\frac{2+\sqrt2}{\sqrt2+1}-\left(\sqrt2+\sqrt3\right)\)
\(=2+\sqrt3+\frac{\sqrt2\left(\sqrt2+1\right)}{\sqrt2+1}-\sqrt2-\sqrt3\)
\(=2-\sqrt2+\sqrt2\)
=2
bài 1:
\(a.\sqrt{25 . 144}=\sqrt{25}.\sqrt{144}=5.12=60\)
\(b.\sqrt{45 . 80}=\sqrt{9 . 5 . 5 . 16}=\sqrt{9 . 25 . 16}=\sqrt{9}.\sqrt{25}.\sqrt{16}=3.5.4=60\)
\(c.\sqrt{52}.\sqrt{13}=\sqrt{52 . 13}=\sqrt{4 . 13 . 13}=\sqrt{4 . 13^2}=\sqrt{4}.\sqrt{13^2}=2.13=26\)
\(d.\sqrt{7}.\sqrt{28}=\sqrt{7 . 28}=\sqrt{7 . 7 . 4}=\sqrt{7^2 . 4}=\sqrt{7^2}.\sqrt{4}=7.2=14\)
\(e.\sqrt{1 \frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{\sqrt{25}}{\sqrt{16}}=\frac{5}{4}\)
\(f.\sqrt{\frac{25}{64}}=\frac{\sqrt{25}}{\sqrt{64}}=\frac{5}{8}\)
\(g.\frac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\frac{12,5}{0,5}}=\sqrt{25}=5\)
\(h.\frac{\sqrt{230}}{\sqrt{2,3}}=\sqrt{\frac{230}{2,3}}=\sqrt{100}=10\)
bài 2:
\(a.\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{50}{3}}-\sqrt{24}\right).\sqrt{6}\)
\(= \sqrt{\frac{2}{3}} . \sqrt{6} + \sqrt{\frac{50}{3}} . \sqrt{6} - \sqrt{24} . \sqrt{6}\)
\(= \sqrt{\frac{2}{3} . 6} + \sqrt{\frac{50}{3} . 6} - \sqrt{24 . 6}\)
\(= \sqrt{4} + \sqrt{100} - \sqrt{144}\)
\(=2+10-12=0\)
b. \(\sqrt{3 + \sqrt{5}}.\sqrt{2}=\sqrt{(3 + \sqrt{5}) . 2}\)
\(=\sqrt{6 + 2\sqrt{5}}=\sqrt{5 + 2\sqrt{5} + 1}\)
\(=\sqrt{(\sqrt{5} + 1)^2}=\vert{}\sqrt{5}+1\vert{}=\sqrt{5}+1\)
\(c.\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\sqrt{\frac{4}{3}}\right).\sqrt{12}\)
\(= \sqrt{\frac{3}{4}} . \sqrt{12} - \sqrt{3} . \sqrt{12} + 5\sqrt{\frac{4}{3}} . \sqrt{12}\)
\(= \sqrt{\frac{3}{4} . 12} - \sqrt{3 . 12} + 5\sqrt{\frac{4}{3} . 12}\)
\(= \sqrt{9} - \sqrt{36} + 5\sqrt{16}\)
\(= 3 - 6 + 5 . 4\)
\(=3-6+20=17\)
\(d.\sqrt{3 - \sqrt{5}}.\sqrt{8}=\sqrt{3 - \sqrt{5}}.\sqrt{2}.\sqrt{4}\)
\(=\sqrt{(3 - \sqrt{5}) . 2}.2=2\sqrt{6 - 2\sqrt{5}}\)
\(=2\sqrt{5 - 2\sqrt{5} + 1}=2\sqrt{(\sqrt{5} - 1)^2}\)
\(=2\vert{}\sqrt{5}-1\vert{}=2(\sqrt{5}-1)=2\sqrt{5}-2\)
bài 3:
\(a.\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)
\(= \sqrt{\frac{1}{7}} : \sqrt{7} - \sqrt{\frac{16}{7}} : \sqrt{7} + \sqrt{7} : \sqrt{7}\)
\(= \sqrt{\frac{1}{7} : 7} - \sqrt{\frac{16}{7} : 7} + 1\)
\(= \sqrt{\frac{1}{49}} - \sqrt{\frac{16}{49}} + 1\)
\(=\frac{1}{7}-\frac{4}{7}+1=-\frac{3}{7}+1=\frac{4}{7}\)
\(b.\sqrt{36 - 12\sqrt{5}}:\sqrt{6}=\sqrt{\frac{36 - 12\sqrt{5}}{6}}\)
\(=\sqrt{6 - 2\sqrt{5}}=\sqrt{5 - 2\sqrt{5} + 1}\)
\(=\sqrt{(\sqrt{5} - 1)^2}=\vert{}\sqrt{5}-1\vert{}=\sqrt{5}-1\)
\(c.\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(= \sqrt{\frac{1}{3}} : \sqrt{3} - \sqrt{\frac{4}{3}} : \sqrt{3} + \sqrt{3} : \sqrt{3}\)
\(= \sqrt{\frac{1}{3} : 3} - \sqrt{\frac{4}{3} : 3} + 1\)
\(=\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+1=\frac{1}{3}-\frac{2}{3}+1\)
\(=-\frac{1}{3}+1=\frac{2}{3}\)
\(e.\sqrt{3 - \sqrt{5}}:\sqrt{2}=\sqrt{\frac{3 - \sqrt{5}}{2}}\)
\(=\sqrt{\frac{6 - 2\sqrt{5}}{4}}=\frac{\sqrt{6 - 2\sqrt{5}}}{\sqrt{4}}\)
\(=\frac{\sqrt{5 - 2\sqrt{5} + 1}}{2}=\frac{\sqrt{(\sqrt{5} - 1)^2}}{2}\)
\(=\frac{\vert{}\sqrt{5} - 1\vert{}}{2}=\frac{\sqrt{5} - 1}{2}\)
bài 4:
\(a.\sqrt{1,6}.\sqrt{250}+\sqrt{19,6}:\sqrt{4,9}=\sqrt{1,6 . 250}+\sqrt{\frac{19,6}{4,9}}\)
\(=\sqrt{400}+\sqrt{4}=20+2=22\)
\(b.\sqrt{1 \frac{3}{4}}.\sqrt{2 \frac{2}{7}}.\sqrt{5 \frac{4}{9}}=\sqrt{\frac{7}{4}}.\sqrt{\frac{16}{7}}.\sqrt{\frac{49}{9}}\)
\(=\sqrt{\frac{7}{4} . \frac{16}{7} . \frac{49}{9}}=\sqrt{\frac{16 . 49}{4 . 9}}=\sqrt{\frac{4 . 49}{9}}\)
\(=\frac{\sqrt{4} . \sqrt{49}}{\sqrt{9}}=\frac{2 . 7}{3}=\frac{14}{3}\)
\(c.\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)
\(= 20\sqrt{300} : \sqrt{15} - 15\sqrt{675} : \sqrt{15} + 5\sqrt{75} : \sqrt{15}\)
\(= 20\sqrt{\frac{300}{15}} - 15\sqrt{\frac{675}{15}} + 5\sqrt{\frac{75}{15}}\)
\(= 20\sqrt{20} - 15\sqrt{45} + 5\sqrt{5}\)
\(= 20\sqrt{4 . 5} - 15\sqrt{9 . 5} + 5\sqrt{5}\)
\(= 20 . 2\sqrt{5} - 15 . 3\sqrt{5} + 5\sqrt{5}\)
\(= 40\sqrt{5} - 45\sqrt{5} + 5\sqrt{5}\)
\(= (40 - 45 + 5)\sqrt{5}\)
\(=0\sqrt{5}=0\)
d. \(\left( \sqrt{325} - \sqrt{117} + 2\sqrt{208} \right) : \sqrt{13}\)
\(= \sqrt{325} : \sqrt{13} - \sqrt{117} : \sqrt{13} + 2\sqrt{208} : \sqrt{13}\)
\(= \sqrt{\frac{325}{13}} - \sqrt{\frac{117}{13}} + 2\sqrt{\frac{208}{13}}\)
\(= \sqrt{25} - \sqrt{9} + 2\sqrt{16}\)
\(=5-3+2.4=10\)
\(e.\frac{2\sqrt{8} - \sqrt{12}}{\sqrt{18} - \sqrt{48}}.\frac{\sqrt{5} + \sqrt{27}}{\sqrt{30} + \sqrt{162}}=\frac{2\sqrt{4 . 2} - \sqrt{4 . 3}}{\sqrt{9 . 2} - \sqrt{16 . 3}}.\frac{\sqrt{5} + \sqrt{27}}{\sqrt{6 . 5} + \sqrt{81 . 2}}\)
\(=\frac{2 . 2\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} - 4\sqrt{3}}.\frac{\sqrt{5} + 3\sqrt{3}}{\sqrt{6}.\sqrt{5} + 9\sqrt{2}}=\frac{4\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} - 4\sqrt{3}}.\frac{\sqrt{5} + 3\sqrt{3}}{\sqrt{6}(\sqrt{5} + 3\sqrt{3})}\)
\(=\frac{2(2\sqrt{2} - \sqrt{3})}{3\sqrt{2} - 4\sqrt{3}}.\frac{1}{\sqrt{6}}=\frac{2(2\sqrt{2} - \sqrt{3})}{\sqrt{6}(3\sqrt{2} - 4\sqrt{3})}\)
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