Toán

a, Đặt \(A=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)

\(\Rightarrow2A=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)

\(\Rightarrow2A-A=2^{2011}-2^0\)

\(\Rightarrow A=2^{2011}-1\)

b,\(7^{x+2}+2.7^{x-1}=345\)

\(7^{x-1}.\left(7^3+2\right)=345\)

\(\Rightarrow7^{x-1}.345=345\)

\(\Rightarrow7^{x-1}=345:345=1\)

\(\Rightarrow7^{x-1}=7^0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

Vậy \(x=1\)

a, Vẽ phân giác AD của góc BAC

Kẻ BH\(\perp\)AD tại H ; CK\(\perp AD\) tại K

Dễ thấy \(sin\widehat{A_1}=sin\widehat{A_2}=sin\dfrac{A}{2}=\dfrac{BH}{AB}=\dfrac{CK}{AC}=\dfrac{BH+CK}{AB+AC}\le\)\(\le\dfrac{BD+CD}{b+c}=\dfrac{a}{b+c}\)

b, Tượng tự \(sin\dfrac{B}{2}\le\dfrac{b}{a+c};sin\dfrac{C}{2}\le\dfrac{c}{a+b}\)

Mặt khác \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc\)

\(\Rightarrow sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2}\le\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)

Hình tự vẽ nhé bạn

Áp dụng định lý Py-ta-go trong tam giác vuông ABC ta có:

AB²+AC²=BC²

⇔27²+36²=BC²

⇒BC=45

Ta có: Sin \(\widehat{ABC}\) = \(\dfrac{AC}{BC}=\dfrac{36}{45}\)

⇒ \(\widehat{ABC}\) ≃ 53^o8^'

⇒ \(\widehat{ACB}=90^0-\widehat{ABC}\) = 36^o52^'

a,Ta có: x- 1/2= y- 2/3⇒ y= x- 1/2 +2/3= x+ 1/6

x- 1/2= z- 3/4⇒ z= x+ 1/4

⇒ x- 2y+3z= 24

⇔ x- 2( x+ 1/6)+ 3( x+1/4)= 24

⇔ x- 2x- 1/3+ 3x+ 3/4= 24

⇔ 2x= 283/12

⇔ x= 283/24

b, Ta có: xy= -30⇒ x= -30/y

yz= 42⇒ z= 42/y

lại có: z- x= -12

⇒ 42/y+ 30/y= -12

⇔ 72/y= -12

⇒ y= -6

⇒ x= 5; z= -7

Ta co :x+y=x*y =>(x.y)-(x+y)=0 =>x.y-x-y+1=1 =>x.(y-1)-(y-1)=1 =>(y-1).(x-1)=1 =>y-1=x-1=1 hoac y-1=x-1=-1 =>y=x=2 hoac y=x=0 Vay y=x=2 hoac y=x=0

mình ko bít giải chỉ bít kết quả thui

2+2=2.2

Suy ra x=2 và y=2

\(a,\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)

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

\(=2\)

\(b,\sqrt{\left(\sqrt{5}-\sqrt{10}\right)^2}-\sqrt{10.\left(\sqrt{2+1}\right)^2}\)

\(=\left|\sqrt{5}-\sqrt{10}\right|-\sqrt{10.3}\)

\(=\left(\sqrt{10}-\sqrt{5}\right)-\sqrt{30}\)

\(=\sqrt{10}-\sqrt{5}-\sqrt{30}\)

\(=\sqrt{5}.\left(\sqrt{2}-1-\sqrt{6}\right)\)

\(c,\left(\sqrt{2}+\sqrt{3}\right)^2.\sqrt{49-20\sqrt{6}}\)

\(=\left(2+3+2.\sqrt{2}.\sqrt{3}\right).\sqrt{\left(5-2\sqrt{6}\right)^2}\)

\(=\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)\)

\(=1\)

d,\(\dfrac{2}{\sqrt{8-\sqrt{60}}}-\sqrt{\dfrac{\sqrt{18}+\sqrt{27}}{\sqrt{3}+\sqrt{2}}}\)

\(=\dfrac{2}{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}-\sqrt{\dfrac{3\sqrt{2}+3\sqrt{3}}{\sqrt{3}+\sqrt{2}}}\)

\(=\dfrac{2}{\sqrt{5}-\sqrt{3}}-\sqrt{\dfrac{3.\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{3}+\sqrt{2}}}\)

\(=\dfrac{2}{\sqrt{5}-\sqrt{3}}-\sqrt{3}\)

\(=\dfrac{2.\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\sqrt{3}\)

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

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

\(=\dfrac{2\sqrt{5}}{2}\)

\(=\sqrt{5}\)

4x⁴+4x²y²-8y⁴

=4x⁴-4x²y²+8x²y²-8y⁴

=4x²(x²-y²)+8y²(x²-y²)

=(x²-y²)(4x²-8y²)

=4(x²-y²)(x²-2y²)

bạn ơi ghi lại đề rõ dùm mình đi

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\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)

\(=[1+\frac{\sqrt{a}.\left(\sqrt{a}+1\right)}{\sqrt{a+1}}].[1-\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{\sqrt{a}-1}]\)

\(=\left(1+\sqrt{a}\right).\left(1-\sqrt{a}\right)\)

\(=1-a\)

\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\\ =\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\\ =\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\\ =a-b\)

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