\(\dfrac{6}{5}x=\dfrac{9}{2}y=\dfrac{18}{7}z\)

=>\(\dfrac{x}{\dfrac{5}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{7}{18}}\)

Đặt \(\dfrac{x}{\dfrac{5}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{7}{18}}=k\)

=>\(x=\dfrac{5}{6}k;y=\dfrac{2}{9}k;z=\dfrac{7}{18}k\)

\(2x^2+3y^2-z^2=4\)

=>\(2\cdot\left(\dfrac{5}{6}k\right)^2+3\cdot\left(\dfrac{2}{9}k\right)^2-\left(\dfrac{7}{18}k\right)^2=4\)

=>\(\dfrac{50}{36}k^2+\dfrac{4}{27}k^2-\dfrac{49}{324}k^2=4\)

=>\(k^2=\dfrac{1296}{449}\)

=>\(k=\pm\dfrac{36}{\sqrt{449}}\)

TH1: \(k=\dfrac{36}{\sqrt{449}}\)

=>\(x=\dfrac{5}{6}\cdot\dfrac{36}{\sqrt{449}}=\dfrac{30}{\sqrt{449}};y=\dfrac{2}{9}\cdot\dfrac{36}{\sqrt{449}}=\dfrac{8}{\sqrt{449}};z=\dfrac{7}{18}\cdot\dfrac{36}{\sqrt{449}}=\dfrac{14}{\sqrt{449}}\)

TH2: \(k=-\dfrac{36}{\sqrt{449}}\)

=>\(x=\dfrac{5}{6}\cdot\dfrac{-36}{\sqrt{449}}=\dfrac{-30}{\sqrt{449}};y=\dfrac{2}{9}\cdot\dfrac{-36}{\sqrt{449}}=\dfrac{-8}{\sqrt{449}};z=\dfrac{7}{18}\cdot\dfrac{-36}{\sqrt{449}}=\dfrac{-14}{\sqrt{449}}\)

774 ha

`90%` của `860` ha là: `860 \times 0,9 = 774.`

a: Các tia đối nhau là Ax và Ay; Ax và AB

Các tia trùng nhau là Ay và AB

b: Hai tia không có điểm chung là By và Ax

Thời gian chạy của Lan là \(8p=\dfrac{2}{15}\left(phút\right)\)

\(\dfrac{x}{y}=\dfrac{5}{3}\)

=>\(\dfrac{x}{5}=\dfrac{y}{3}\)

mà x+2y=4,4

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x+2y}{5+2\cdot3}=\dfrac{4.4}{11}=0,4\)

=>\(x=0,4\cdot5=2;y=0,4\cdot3=1,2\)

`#3107.101107`

\(\dfrac{x}{y}=\dfrac{5}{3}\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{5}=\dfrac{2y}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{5}=\dfrac{2y}{6}=\dfrac{x+2y}{5+6}=\dfrac{4,4}{11}=0,4\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=0,4\)

\(\Rightarrow x=0,4\cdot5=2\) `;` \(y=0,4\cdot3=1,2\)

Vậy, `x = 2; y = 1,2.`

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2024}}\)

=>\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2023}}\)

=>\(2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2023}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{2024}}\)

=>\(A=1-\dfrac{1}{2^{2024}}\)

\(223-x\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2024}}\right):\left(1-\dfrac{1}{2^{2024}}\right)=23\)

=>\(223-x\left(1-\dfrac{1}{2^{2024}}\right):\left(1-\dfrac{1}{2^{2024}}\right)=23\)

=>223-x=23

=>x=200

2x=3y

=>\(\dfrac{x}{3}=\dfrac{y}{2}\)

=>\(\dfrac{x}{21}=\dfrac{y}{14}\left(1\right)\)

5y=7z

=>\(\dfrac{y}{7}=\dfrac{z}{5}\)

=>\(\dfrac{y}{14}=\dfrac{z}{10}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)

Đặt \(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=k\)

=>x=21k; y=14k; z=10k

3x-2y+5z=-30

=>\(3\cdot21k-2\cdot14k+5\cdot10k=-30\)

=>85k=-30

=>\(k=-\dfrac{30}{85}=-\dfrac{6}{17}\)

=>\(x=21\cdot\dfrac{-6}{17}=\dfrac{-126}{17};y=14\cdot\dfrac{-6}{17}=-\dfrac{84}{17};z=10\cdot\dfrac{-6}{17}=-\dfrac{60}{17}\)

\(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\)

\(5y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\)

\(\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x}{63}=\dfrac{2y}{28}=\dfrac{5z}{50}=\dfrac{3x-2y+5z}{63-28+50}=\dfrac{-30}{85}\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{30}{85}.21=-\dfrac{126}{17}\\y=-\dfrac{30}{85}.14=-\dfrac{84}{17}\\z=-\dfrac{30}{85}.10=-\dfrac{60}{17}\end{matrix}\right.\)

Em có ghi nhầm đề đâu ko mà kết quả xấu quá

\(A=\dfrac{3^{2022}+2}{3^{2022}-1}=\dfrac{3^{2022}-1+3}{3^{2022}-1}=1+\dfrac{3}{3^{2022}-1}\)

\(B=\dfrac{3^{2022}}{3^{2022}-3}=\dfrac{3^{2022}-3+3}{3^{2022}-3}=1+\dfrac{3}{3^{2022}-3}\)

Vì \(3^{2022}-1>3^{2022}-3\)

nên \(\dfrac{3}{3^{2022}-1}< \dfrac{3}{3^{2022}-3}\)

=>\(1+\dfrac{3}{3^{2022}-1}< 1+\dfrac{3}{2^{2022}-3}\)

=>A<B

Với các số dương \(a;b;n\) sao cho \(a>b\) ta luôn có: \(\dfrac{a}{b}>\dfrac{a+n}{b+n}\)

Thật vậy, do \(a>b\Rightarrow an>bn\Rightarrow ab+an>ab+bn\)

\(\Rightarrow a\left(b+n\right)>b\left(a+n\right)\)

\(\Rightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\)

Áp dụng:

 Do \(3^{2022}>3^{2022}-3>0\) và \(2>0\) nên:

\(\dfrac{3^{2022}}{3^{2022}-3}>\dfrac{3^{2022}+2}{3^{2022}-3+2}\Rightarrow\dfrac{3^{2022}}{3^{2022}-3}>\dfrac{3^{2022}+2}{3^{2022}-1}\)

Vậy \(B>A\)

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