Bài 1: Căn bậc hai

\(P=\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{77}+\sqrt{79}}+\dfrac{1}{\sqrt{79}+\sqrt{81}}\)

\(P=\dfrac{\sqrt{1}-\sqrt{3}}{\left(\sqrt{1}+\sqrt{3}\right)\left(\sqrt{1}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}+...+\dfrac{\sqrt{79}-\sqrt{81}}{\left(\sqrt{79}+\sqrt{81}\right)\left(\sqrt{79}-\sqrt{81}\right)}\)

\(P=\dfrac{\sqrt{1}-\sqrt{3}}{1-3}+\dfrac{\sqrt{3}-\sqrt{5}}{3-5}+...+\dfrac{\sqrt{79}-\sqrt{81}}{79-81}\)

\(P=\dfrac{\sqrt{1}-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}-\sqrt{7}+...+\sqrt{77}-\sqrt{79}+\sqrt{79}-\sqrt{81}}{-2}\)

\(P=\dfrac{1-9}{-2}\)

\(P=\dfrac{-8}{-2}\)

\(P=4\)

⇒ Chọn B

\(P=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+...+\dfrac{\sqrt{81}-\sqrt{79}}{2}\)

\(=\dfrac{-1+\sqrt{3}-\sqrt{3}+\sqrt{5}-...-\sqrt{79}+9}{2}=\dfrac{8}{2}=4\)

a) \(\sqrt{4x+20}-2\sqrt{x+5}+\sqrt{9x+45}=6\) (ĐK: \(x\ge-5\))

\(\Leftrightarrow2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow\sqrt{x+5}=\dfrac{6}{3}\)

\(\Leftrightarrow\sqrt{x+5}=2\)

\(\Leftrightarrow x+5=2^2\)

\(\Leftrightarrow x+5=4\)

\(\Leftrightarrow x=4-5\)

\(\Leftrightarrow x=-1\)

b) \(\sqrt{9x^2-6x+1}=9\)

\(\Leftrightarrow\sqrt{\left(3x-1\right)^2}=9\)

\(\Leftrightarrow\left(3x-1\right)^2=9^2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=9\\3x-1=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=10\\3x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{8}{3}\end{matrix}\right.\)

c) \(\sqrt{2x-1}-2\sqrt{x}+1=0\) (ĐK: \(x\ge\dfrac{1}{2}\))

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

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

\(\Leftrightarrow2x-1=4x-4\sqrt{x}+1\)

\(\Leftrightarrow4x-2x-4\sqrt{x}=-1-1\)

\(\Leftrightarrow2x-4\sqrt{x}=-2\)

\(\Leftrightarrow2x-4\sqrt{x}+2=0\)

\(\Leftrightarrow2\left(x-2\sqrt{x}+1\right)=0\)

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

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

\(\Leftrightarrow\sqrt{x}=1\)

\(\Leftrightarrow x=1\)

a) ĐK: x>=-5

pt <=> \(\sqrt{4\left(x+5\right)}-2\sqrt{x+5}+\sqrt{9\left(x+5\right)}=6\)

\(\Leftrightarrow2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow\sqrt{x+5}=2\)

\(\Leftrightarrow x+5=4\)

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy, pt có nghiệm duy nhất là x=-1

b) ĐK: x thuộc R

pt<=> \(\sqrt{\left(3x-1\right)^2}=9\)

\(\Leftrightarrow\left|3x-1\right|=9\Leftrightarrow\left[{}\begin{matrix}3x-1=9\\3x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{8}{3}\end{matrix}\right.\)

Vậy, pt có tập nghiệm là \(S=\left\{\dfrac{10}{3};-\dfrac{8}{3}\right\}\)

c) ĐK: \(x\ge\dfrac{1}{2}\)

pt<=> \(\sqrt{2x-1}=2\sqrt{x}-1\)

\(\Leftrightarrow2x-1=4x+1-4\sqrt{x}\)

\(\Leftrightarrow2x+2=4\sqrt{x}\)

\(\Leftrightarrow x+1=2\sqrt{x}\)

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

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

Vậy, pt có nghiệm duy nhất là x=1

a: góc B=90-50=40 độ

Xét ΔABC vuông tại A có sin B=AC/BC

=>AC/13=sin40

=>\(AC\simeq8,36\left(cm\right)\)

=>\(AB\simeq9,96\left(cm\right)\)

b: AB*cosB+AC*cosC

=AB*AB/BC+AC*AC/BC

=(AB^2+AC^2)/BC=BC

2)

Gọi tam giác tạo thành là tam giác MNP: 

Ta có: 

\(tgN=\dfrac{MP}{NP}\Rightarrow tg50^o=\dfrac{MP}{7,5}\Rightarrow MP=tg50^o\cdot7,5\approx8,94\left(m\right)\)

Vậy cây cột đèn cao 8,94 m

1) \(1\left(p\right)=\dfrac{1}{60}\left(h\right)\)

Sau 1 phút thì máy bay bay được số km là (AB) :

\(750\cdot\dfrac{1}{60}=12,5\left(km\right)\)

Sau 1 phút thì máy bay bay cao được số km là (BH)

Ta có: \(sinA=\dfrac{BH}{AB}\Rightarrow sin35^o=\dfrac{BH}{12,5}\Rightarrow BH=sin35^o\cdot12,5\approx7,17\left(km\right)\)

a: góc C=90-38=52 độ

Xét ΔABC vuông tại A có sin B=AC/BC

=>4/BC=sin38

=>\(BC\simeq6,50\left(cm\right)\)

=>\(AB\simeq5,12\left(cm\right)\)

b: \(BC=\sqrt{5^2+7^2}=\sqrt{74}\simeq8,6\left(cm\right)\)

tan B=AC/AB=5/7

=>góc B\(\simeq36^0\)

=>\(\widehat{C}\simeq54^0\)

c: góc B=90-43=47 độ

Xét ΔABC vuông tại A có

sin C=AB/BC

=>AB/8=sin43

=>\(AB\simeq5,46\left(cm\right)\)

=>\(AC\simeq5,85\left(cm\right)\)

\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}\)

\(=\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}\)

\(=2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}\)

\(=2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}\)

\(=2\sqrt{x-5}\)

ĐK: x>=5

 \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}\)

\(=2\sqrt{x-5}\)

\(VT=\left(\dfrac{\sqrt{14}.\sqrt{14}}{\sqrt{14}}+\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{5}}\right)\sqrt{5-\sqrt{21}}\\ =\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\\ =\sqrt{14}.\sqrt{5-\sqrt{2}1}+\sqrt{6}.\sqrt{5-\sqrt{21}}\\ =\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\\ =\sqrt{49-2.7.\sqrt{21}+21}+\sqrt{9-2.3.\sqrt{21}+21}\\ =\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(3-\sqrt{21}\right)^2}\\ =\left|7-\sqrt{21}\right|+\left|3-\sqrt{21}\right|\\ =7-\sqrt{21}+\sqrt{21}-3\\ =7-3=4=VP\)

help me now

\(VT=\left(\sqrt{14}+\sqrt{6}\right)\cdot\sqrt{5-\sqrt{21}}\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\cdot\sqrt{10-2\sqrt{21}}\)

=(căn 7+căn 3)(căn 7-căn 3)

=7-3

=4=VP

\(\dfrac{4}{\sqrt{7}+\sqrt{3}}+\dfrac{4}{\sqrt{7}-\sqrt{3}}\\ =\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}+\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}\\ =\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{7-3}+\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{7-3}\\ =\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{4}+\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{4}\\ =\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\\ =2\sqrt{7}\)

@seven

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

=2*căn 7

1)

`2(x-1)^2 =32`

`<=>(x-1)^2 =18`

`<=>x-1=9` hoặc `x-1=-9`

`<=>x=10` hoặc `x=-8`

2)

`(x+1)(81x^2 -9)=0`

`<=>(x+1)(9x-3)(9x+3)=0`

`<=>x+1=0` hoặc `9x-3=0` hoặc `9x+3=0`

`<=>x=-1` hoặc `x=1/3` hoặc `x=-1/3`