Giúp mình với mọi người ai làm đc câu nào thì làm :)
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NV
1
Những câu hỏi liên quan
NN
1
NN
1
12 tháng 10 2021
a, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
b, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{2021a}{2021b}=\dfrac{2021a-c}{2021b-d}\)
c, Ta có \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a}{b}\right)^2=\left(\dfrac{c}{d}\right)^2\)
Áp dụng t/c dtsbn:
\(\left(\dfrac{a}{b}\right)^2=\left(\dfrac{c}{d}\right)^2=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\)
NN
0
NN
2
1 tháng 11 2021
Câu 20:
Ta có: \(\widehat{A}-\widehat{B}=40^0\Rightarrow\widehat{B}=\widehat{A}-40^0\)
\(\widehat{A}=2\widehat{C}\Rightarrow\widehat{C}=\frac{\widehat{A}}{2}\)
Vì AB//CD (gt) \(\Rightarrow\widehat{A}+\widehat{D}=180^0\)(hai góc trong cùng phía)\(\Rightarrow\widehat{D}=180^0-\widehat{A}\)
Tứ giác ABCD \(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\Rightarrow\widehat{A}+\left(\widehat{A}-40^0\right)+\frac{\widehat{A}}{2}+\left(180^0-\widehat{A}\right)=360^0\)
Và đến đây bạn dễ dàng tìm được góc A và từ đó suy ra được góc D.
1 tháng 11 2021
Câu 29: Ta có:
\(\hept{\begin{cases}xy+x+y=3\\yz+y+z=8\\xz+x+z=15\end{cases}}\Leftrightarrow\hept{\begin{cases}xy+x+y+1=4\\yz+y+z+1=9\\xz+x+z+1=16\end{cases}\Leftrightarrow}\hept{\begin{cases}x\left(y+1\right)+\left(y+1\right)=4\\y\left(z+1\right)+\left(z+1\right)=9\\x\left(z+1\right)+\left(z+1\right)=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)\left(y+1\right)=4\\\left(y+1\right)\left(z+1\right)=9\\\left(z+1\right)\left(x+1\right)=16\end{cases}}\)
Đặt \(\hept{\begin{cases}x+1=a\\y+1=b\\z+1=c\end{cases}}\)với a,b,c > 1, khi đó ta có
\(\hept{\begin{cases}ab=4\\bc=9\\ca=16\end{cases}}\Leftrightarrow\hept{\begin{cases}abbc=4.9\\c=\frac{9}{b}\\ca=16\end{cases}}\Leftrightarrow\hept{\begin{cases}16b^2=36\\c=\frac{9}{b}\\a=\frac{16}{c}\end{cases}}\Leftrightarrow\hept{\begin{cases}b^2=\frac{36}{16}=\frac{9}{4}\\c=\frac{9}{b}\\a=\frac{16}{c}\end{cases}}\Leftrightarrow\hept{\begin{cases}b=\frac{3}{2}\\c=\frac{9}{\frac{3}{2}}=6\\a=\frac{16}{6}=\frac{8}{3}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=a-1=\frac{8}{3}-1=\frac{5}{3}\\y=b-1=\frac{3}{2}-1=\frac{1}{2}\\z=c-1=6-1=5\end{cases}}\)
Vậy \(P=x+y+z=\frac{5}{3}+\frac{1}{2}+5=\frac{10+3+30}{6}=\frac{43}{6}\)
NN
1
20 tháng 8 2021
Bài 5:
Ta có : \(\widehat{A_1}+\widehat{A_3}=180^o\) (kề bù)
\(100^o+\widehat{A_3}=180^o\)
\(\widehat{A_3}=80^o\)
Ta có: \(\widehat{A_3}=\widehat{B_1}=80^o\)
\(\widehat{A_3}\) và \(\widehat{B_1}\) ở vị trí đồng vị
\(\Rightarrow AC//BD\)
\(\Rightarrow\widehat{C}_1=\widehat{D_1}=135^o\) (đồng vị)
\(x=135^o\)
b)
Ta có: \(\widehat{G_1}+\widehat{B_1}=180^o\left(120^o+60^o=180^o\right)\)
\(\widehat{G_1}\) và \(\widehat{B_1}\) ở vị trí trong cùng phía
\(\Rightarrow QH//BK\)
\(\Rightarrow\widehat{H_1}=\widehat{K_1}=90^o\)(so le)
\(x=90^o\)
NN
0
K
13 tháng 12 2020
11 c)
\(a^2+2\ge2\sqrt{a^2+1}\Leftrightarrow a^2+1-2\sqrt{a^2+1}+1\ge0\Leftrightarrow\left(\sqrt{a^2+1}-1\right)^2\ge0\) (luôn đúng)
K
13 tháng 12 2020
12 a) Có a+b+c=1\(\Rightarrow\) (1-a)(1-b)(1-c)= (b+c)(a+c)(a+b) (*)
áp dụng BĐT cô-si: \(\left(b+c\right)\left(a+c\right)\left(a+b\right)\ge2\sqrt{bc}2\sqrt{ac}2\sqrt{ab}=8\sqrt{\left(abc\right)2}=8abc\) ( luôn đúng với mọi a,b,c ko âm )
b) áp dụng BĐT cô-si: \(c\left(a+b\right)\le\dfrac{\left(a+b+c\right)^2}{4}=\dfrac{1}{4}\)
Tương tự: \(a\left(b+c\right)\le\dfrac{1}{4};b\left(c+a\right)\le\dfrac{1}{4}\)
\(\Rightarrow abc\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{4}\dfrac{1}{4}\dfrac{1}{4}=\dfrac{1}{64}\)
P
0
9: \(A=\dfrac{3^2}{10}+\dfrac{3^2}{40}+...+\dfrac{3^2}{340}\)
\(=3\left(\dfrac{3}{10}+\dfrac{3}{40}+...+\dfrac{3}{340}\right)\)
\(=3\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\right)\)
\(=3\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
\(=3\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=3\cdot\dfrac{9}{20}=\dfrac{27}{20}\)
10: \(A=\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{26\cdot31}\)
\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)
\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{25}-\dfrac{1}{31}\right)\)
\(=5\left(1-\dfrac{1}{31}\right)=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)
11: \(A=\dfrac{6}{15}+\dfrac{6}{35}+\dfrac{6}{63}+\dfrac{6}{99}\)
\(=3\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}\right)\)
\(=3\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{11}\right)=3\cdot\dfrac{8}{33}=\dfrac{8}{11}\)
12: \(A=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+...+\dfrac{3}{49\cdot51}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{49\cdot51}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)=\dfrac{3}{2}\cdot\dfrac{16}{51}=\dfrac{8}{17}\)
13: \(A=\dfrac{1}{2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\)
\(=1-\dfrac{1}{16}=\dfrac{15}{16}\)
14: \(A=\dfrac{1}{2}+\dfrac{2}{8}+\dfrac{3}{28}+\dfrac{4}{77}+\dfrac{5}{176}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}\)
15: \(A=\dfrac{3}{54}+\dfrac{5}{126}+\dfrac{7}{294}+\dfrac{8}{609}\)
\(=\dfrac{3}{6\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{7}{14\cdot21}+\dfrac{8}{21\cdot29}\)
\(=\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{29}\)
\(=\dfrac{1}{6}-\dfrac{1}{29}=\dfrac{23}{174}\)
16: \(A=\dfrac{5}{24}+\dfrac{5}{104}+\dfrac{5}{234}+\dfrac{5}{414}\)
\(=\dfrac{5}{3\cdot8}+\dfrac{5}{8\cdot13}+\dfrac{5}{13\cdot18}+\dfrac{5}{18\cdot23}\)
\(=\dfrac{1}{3}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}\)
\(=\dfrac{1}{3}-\dfrac{1}{23}=\dfrac{20}{69}\)
17: \(A=\dfrac{\dfrac{3}{54}+\dfrac{5}{126}+\dfrac{7}{294}}{\dfrac{5}{24}+\dfrac{5}{104}+\dfrac{5}{234}}\)
\(=\dfrac{\dfrac{1}{6}-\dfrac{1}{21}}{\dfrac{1}{3}-\dfrac{1}{18}}=\dfrac{15}{126}:\dfrac{15}{54}=\dfrac{54}{126}=\dfrac{3}{7}\)