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\(a,\sqrt{75}+2\sqrt{3}-2\sqrt{7}\\ =\sqrt{25\cdot3}+2\sqrt{3}-2\sqrt{7}\\ =5\sqrt{3}+2\sqrt{3}-2\sqrt{7}\\ =7\sqrt{3}-2\sqrt{7}\)
\(b,\sqrt{\left(4-\sqrt{7}\right)^2}-\sqrt{63}\\ =\left|4-\sqrt{7}\right|-\sqrt{9\cdot7}\\ =4-\sqrt{7}-3\sqrt{7}\\ =4-4\sqrt{7}\)
\(c,\dfrac{3}{\sqrt{5}+3}-\dfrac{\sqrt{5}}{\sqrt{5}-3}\\ =\dfrac{3\left(\sqrt{5}-3\right)}{5-3}-\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{5-3}\\ =\dfrac{3\sqrt{5}-9-5-3\sqrt{5}}{2}\\ =\dfrac{-14}{2}\\ =-7\)
`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(A=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}\)
\(A=\frac{2}{3}\)
a: Sửa đề: \(\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{3}-2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{3}-2}=\dfrac{2-\sqrt{3}}{\sqrt{3}-2}\)
=-1
b: Sửa đề: \(\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
=1
Lời giải:
$\sqrt{7+2\sqrt{10}}=\sqrt{2+5+2\sqrt{2.5}}=\sqrt{(\sqrt{2}+\sqrt{5})^2}=\sqrt{2}+\sqrt{5}$
\(\sqrt[3]{3\sqrt[3]{3}-3\sqrt[3]{2}-1}=\sqrt[3]{(1-\sqrt[3]{2})^3}=1-\sqrt[3]{2}\)
Do đó:
\(\text{TS}=\sqrt[3]{2}+\sqrt{2}+\sqrt{5}+1-\sqrt[3]{2}=\sqrt{2}+\sqrt{5}+1=\text{MS}\)
\(A=\frac{\text{TS}}{\text{MS}}=1\)
`Answer:`
\(\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(5-7\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|5-7\sqrt{3}\right|\)
\(=2-\sqrt{3}-7\sqrt{3}+5\)
\(=7-8\sqrt{3}\)