Tổng 3 góc của 1 tam giác là 180^o

Hình 1.

90^o + 55^o + x = 180^o

145^o + x = 180^o

x = 180^o - 145^o

x = 35^o

Hình 2.

30^o + x + 40^o = 180^o

70^o + x = 180^o

x = 180^o - 70^o

x = 110^o

Hình 3.

50^o + x + x = 180^o

2x = 180^o - 50^o

2x = 130^o

x = 130^o : 2

x = 65^o

$450-2\times x=75:5-5$

$\Rightarrow 450-2\times x=15-5$

$\Rightarrow 450-2\times x=10$

$\Rightarrow 2\times x=450-10$

$\Rightarrow 2\times x=440$

$\Rightarrow x=440:2$

$\Rightarrow x=220$

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

\(Q=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}}{\dfrac{5}{2021}+\dfrac{5}{2022}-\dfrac{5}{2023}}-\dfrac{\dfrac{2}{2021}+\dfrac{2}{2022}-\dfrac{2}{2023}}{\dfrac{3}{2021}+\dfrac{3}{2022}-\dfrac{3}{2023}}\)

\(=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}}{5\left(\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)}-\dfrac{2\left(\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)}{3\left(\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)}\)

\(=\dfrac{1}{5}-\dfrac{2}{3}=-\dfrac{7}{15}\)

gấp ạ!!

Chiều rộng sân bóng là:

$60\times\frac35=36\text{ }(m)$

Chu vi sân bóng là:

$(60+36)\times2=192\text{ }(m)$

Diện tích sân bóng là:

$60\times36=2160\text{ }(m^2)$

Số tự nhiên lớn nhất có bốn chữ số mà trong đó không có hai chữ số nào giống nhau là: 9876

9876

$\frac{5}{11}\times\frac76-\frac56\times\frac{1}{11}+\frac{6}{11}$

$=\frac{5}{6}\times\frac{7}{11}-\frac56\times\frac{1}{11}+\frac{6}{11}$

$=\frac56\times\left(\frac{7}{11}-\frac{1}{11}\right)+\frac{6}{11}$

$=\frac56\times\frac{6}{11}+\frac{6}{11}$

$=\frac{6}{11}\times\left(\frac56+1\right)$

$=\frac{6}{11}\times\frac{11}{6}=1$

$\frac{5}{11}\times\frac76-\frac56\times\frac{1}{11}+\frac{6}{11}$

$=\frac{5}{6}\times\frac{7}{11}-\frac56\times\frac{1}{11}+\frac{6}{11}$

$=\frac56\times\left(\frac{7}{11}-\frac{1}{11}\right)+\frac{6}{11}$

$=\frac56\times\frac{6}{11}+\frac{6}{11}$

$=\frac{6}{11}\times\left(\frac56+1\right)$

$=\frac{6}{11}\times\frac{11}{6}=1$

Lời giải:

$\frac{131}{171}=1-\frac{40}{171}> 1-\frac{40}{170}=1-\frac{4}{17}=\frac{13}{17}$
----------------------------------

$\frac{51}{61}=1-\frac{10}{61}=1-\frac{100}{610}$

$\frac{515}{616}=1-\frac{101}{616}$

Xét hiệu:

$\frac{100}{610}-\frac{101}{616}=\frac{100.616-101.610}{610.616}$

$=\frac{100(610+6)-101.610}{610.616}$

$=\frac{600-610}{610.616}<0$

$\Rightarrow \frac{100}{610}< \frac{101}{616}$

$\Rightarrow 1-\frac{100}{610}> 1-\frac{101}{616}$

$\Rightarrow \frac{51}{61}> \frac{515}{616}$ 

làm bài nào cx dc.

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:

\(P=-\left(7-4\sqrt{3}\right)+2-\sqrt{3}=-5+3\sqrt{3}\)

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)