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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
$450-2\times x=75:5-5$
$\Rightarrow 450-2\times x=15-5$
$\Rightarrow 450-2\times x=10$
$\Rightarrow 2\times x=450-10$
$\Rightarrow 2\times x=440$
$\Rightarrow x=440:2$
$\Rightarrow x=220$
450 - 2 x X = 15 - 5
450 - 2 x X =10
2 x X = 450 - 10
2 x X = 440
X = 440 : 2
X = 220
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}}{\dfrac{5}{2021}+\dfrac{5}{2022}-\dfrac{5}{2023}}-\dfrac{\dfrac{2}{2021}+\dfrac{2}{2022}-\dfrac{2}{2023}}{\dfrac{3}{2021}+\dfrac{3}{2022}-\dfrac{3}{2023}}\)
\(=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}}{5\left(\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)}-\dfrac{2\left(\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)}{3\left(\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)}\)
\(=\dfrac{1}{5}-\dfrac{2}{3}=-\dfrac{7}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Số tự nhiên lớn nhất có bốn chữ số mà trong đó không có hai chữ số nào giống nhau là: 9876
![](https://rs.olm.vn/images/avt/0.png?1311)
$\frac{5}{11}\times\frac76-\frac56\times\frac{1}{11}+\frac{6}{11}$
$=\frac{5}{6}\times\frac{7}{11}-\frac56\times\frac{1}{11}+\frac{6}{11}$
$=\frac56\times\left(\frac{7}{11}-\frac{1}{11}\right)+\frac{6}{11}$
$=\frac56\times\frac{6}{11}+\frac{6}{11}$
$=\frac{6}{11}\times\left(\frac56+1\right)$
$=\frac{6}{11}\times\frac{11}{6}=1$
$\frac{5}{11}\times\frac76-\frac56\times\frac{1}{11}+\frac{6}{11}$
$=\frac{5}{6}\times\frac{7}{11}-\frac56\times\frac{1}{11}+\frac{6}{11}$
$=\frac56\times\left(\frac{7}{11}-\frac{1}{11}\right)+\frac{6}{11}$
$=\frac56\times\frac{6}{11}+\frac{6}{11}$
$=\frac{6}{11}\times\left(\frac56+1\right)$
$=\frac{6}{11}\times\frac{11}{6}=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\frac{131}{171}=1-\frac{40}{171}> 1-\frac{40}{170}=1-\frac{4}{17}=\frac{13}{17}$
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$\frac{51}{61}=1-\frac{10}{61}=1-\frac{100}{610}$
$\frac{515}{616}=1-\frac{101}{616}$
Xét hiệu:
$\frac{100}{610}-\frac{101}{616}=\frac{100.616-101.610}{610.616}$
$=\frac{100(610+6)-101.610}{610.616}$
$=\frac{600-610}{610.616}<0$
$\Rightarrow \frac{100}{610}< \frac{101}{616}$
$\Rightarrow 1-\frac{100}{610}> 1-\frac{101}{616}$
$\Rightarrow \frac{51}{61}> \frac{515}{616}$
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Với \(x\ge0;x\ne1\):
\(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)
\(=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{2}=\sqrt{x}\left(1-\sqrt{x}\right)=\sqrt{x}-x\)
b, Thay \(x=7-4\sqrt{3}\) vào P, ta được:
\(P=\sqrt{7-4\sqrt{3}}-\left(7-4\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.2+2^2}+4\sqrt{3}-7\)
\(=\sqrt{\left(\sqrt{3}-2\right)^2}+4\sqrt{3}-7\)
\(=\left|\sqrt{3}-2\right|+4\sqrt{3}-7\)
\(=2-\sqrt{3}+4\sqrt{3}-7\) (vì \(\sqrt{3}< 2\))
\(=-5+3\sqrt{3}\)
$Toru$
a) \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\left(x\ge0,x\ne1\right)\\ =\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(x-1\right)^2}{2}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\left(x-1\right)^2}{2}\\ \)
\(=\dfrac{x-2\sqrt{x}+\sqrt{x}-2-\left(x+2\sqrt{x}-\sqrt{x}-2\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\\ =\left[x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)\right].\dfrac{\sqrt{x}-1}{2}\\ \)
\(=-2\sqrt{x}.\dfrac{\sqrt{x}-1}{2}\\ =-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
b) \(x=7-4\sqrt{3}\left(TMDK\right)\)
\(\sqrt{x}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
Thay vào biểu thức P, ta được:
\(P=-\left(7-4\sqrt{3}\right)+2-\sqrt{3}=-5+3\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(E=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\)
\(3E=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)
\(3E-E=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+..+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\right)\)
\(2E=1+\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6E=3+1+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6E-2E=\left(3+1+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
\(4E=3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow E=\dfrac{3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}}{4}=\dfrac{3}{4}-\dfrac{\dfrac{100}{3^{99}}+\dfrac{100}{3^{100}}}{4}< \dfrac{3}{4}\) (đpcm)
Tổng 3 góc của 1 tam giác là 180o
Hình 1.
90o + 55o + x = 180o
145o + x = 180o
x = 180o - 145o
x = 35o
Hình 2.
30o + x + 40o = 180o
70o + x = 180o
x = 180o - 70o
x = 110o
Hình 3.
50o + x + x = 180o
2x = 180o - 50o
2x = 130o
x = 130o : 2
x = 65o