\(B=\dfrac{\sqrt{x}+5+\sqrt{x}-5-10}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(B=\dfrac{2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

B=\(\dfrac{2}{\sqrt{x}+5}\)

b)thay x=50 vào biểu thức

B=\(\dfrac{2}{\sqrt{50}+5}\)=\(\dfrac{2\left(\sqrt{50}-5\right)}{\left(\sqrt{50}+5\right)\left(\sqrt{50}-5\right)}=\dfrac{2\sqrt{50}-10}{25}\)

c)để B>\(\dfrac{1}{6}\)

\(\dfrac{2}{\sqrt{x}+5}>\dfrac{1}{6}\Rightarrow12>\sqrt{x}+5\)

\(\sqrt{x}< 7\Rightarrow x< 49\)

mik lộn chút U=U1=U2 nhé

a)\(\dfrac{1}{Rtđ}=\dfrac{1}{R1}+\dfrac{1}{R2}=\dfrac{1}{8}+\dfrac{1}{12}=\dfrac{5}{24}\)

\(\Rightarrow Rtđ=\)4,8

b)ta có:U=U1+U2=I.Rtđ=3.4,8=14,4

c)I1=\(\dfrac{U1}{R1}=\dfrac{14,4}{8}\)=1,8

I2=I-I1=3-1,8=1,2

a)=\(3\sqrt{5}-8\sqrt{5}+6\sqrt{2}=-5\sqrt{5}+6\sqrt{2}\)

c)=\(\sqrt{8+2\sqrt{6-2\sqrt{5}}}=\sqrt{8+2\sqrt{\left(\sqrt{5}-1\right)^2}}\)

=\(\sqrt{8+2\left(\sqrt{5}-1\right)}=\sqrt{8+2\sqrt{5}-2}\)

=\(\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)

e)=\(\dfrac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\dfrac{16\left(\sqrt{5}+3\right)}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}\)

=\(\dfrac{4\sqrt{5}+4}{4}+\dfrac{3\sqrt{5}+6}{1}+\dfrac{16\sqrt{5}+48}{-4}\)

=\(\sqrt{5}+1+3\sqrt{5}+6-4\sqrt{5}-12\)=-5

bài 5

a) áp dụng pytago vào \(\Delta ABH\) vuông tại H

\(AH=\sqrt{AB^2-BH^2}=\sqrt{8^2-6^2}=2\sqrt{7}\)

b)ta có:\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\Rightarrow\dfrac{1}{AC^2}=\dfrac{1}{AH^2}-\dfrac{1}{AB^2}\)

=\(\dfrac{1}{\left(2\sqrt{7}\right)^2}-\dfrac{1}{8^2}=\dfrac{9}{448}\)

\(\Rightarrow AC=7,1\)

c)áp dụng pytago vào \(\Delta ABC\) vuông tại A

\(BC=\sqrt{AB^2+AC^2}=\sqrt{8^2+7,1^2}=10,7\)

HC=BC-BH=10,7-6=4,7

bạn tự vẽ hình giúp mik nha

áp dụng định lý đảo pytago vào \(\Delta ABC\)

ta có:\(AB^2+AC^2=10,7^2+6,5^2=156,74\)

\(BC^2=12,5^2=156,25\)

\(\Rightarrow\)\(\Delta ABC\) là tam giác vuông tại A

gọi AH là đườngcao

ta có:AH.BC=AB.AC\(\Rightarrow AH=\dfrac{AB.AC}{BC}\left(đlý\right)=\dfrac{10,7.6,5}{12,5}=5,564\)

b)\(BH=\sqrt{AB^2-AH^2}\left(pytago\right)=\sqrt{10,7^2-5,564^2}=9,1\)

\(HC=\sqrt{AC^2-AH^2}\left(pytago\right)=\sqrt{6,5^2-5,564^2}=3,4\)

a)=\(sin^243+sin^244+sin^245+cos^244+cos^243\)

=\(2+sin^245=\dfrac{5}{2}\)

b)=\(cos^21+cos^22+...+cos^245+sin^244+sin^243+...+sin^21\)

=\(\left(cos^21+sin^21\right)+\left(cos^22+sin^22\right)+...+cos^245\)

=44+\(cos^245\)=44,5

c)=\(\sqrt{\dfrac{4cot68}{cot68}}+1=\sqrt{4}+1=2+1=3\)

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