Bài 7: Tỉ lệ thức

\(\dfrac{1}{3}< \dfrac{1}{1.2}\\ \dfrac{1}{3^2}< \dfrac{1}{2.3}\\ \dfrac{1}{3^3}< \dfrac{1}{3.4}\\ ......\\ \dfrac{1}{3^{99}}< \dfrac{1}{99.100}\)

cộng vế với vế, ta được :\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+......+\dfrac{1}{3^{99}}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)

Ta có: \(\dfrac{x}{y.z}.\dfrac{y}{z.x}=\dfrac{xy}{xyz^2}=\dfrac{1}{z^2}\)

\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\)

\(\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\)

\(\Leftrightarrow6x-3y=2x+2y\)

\(\Leftrightarrow6x-2x=2y+3y\)

\(\Leftrightarrow4x=5y\)

\(\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)

Vậy ..

Ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)

\(\Rightarrow6x-3y=2x+2y\)

\(\Rightarrow6x-2x=2y+3y\)

\(\Rightarrow4x=5y\Rightarrow\dfrac{x}{y}=\dfrac{5}{4}\)

Vậy ...

1/ Ta có: -2x = 5y \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{-2}\)

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

\(\dfrac{x}{5}=\dfrac{y}{-2}=\dfrac{x+y}{5+\left(-2\right)}=\dfrac{30}{3}=10\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=10\\\dfrac{y}{-2}=10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.\left(-2\right)-20\end{matrix}\right.\)

Vậy x = 50; y = -20.

2/ Ta có: 3x = 5y \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\)

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

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x+y}{5+3}=\dfrac{40}{8}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=5\\\dfrac{y}{3}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=5.3=15\end{matrix}\right.\)

Vậy x = 25; y = 15.

3/ Ta có: 4x = 5y \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}\)

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

\(\dfrac{3x}{15}=\dfrac{2y}{8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=5\\\dfrac{y}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=5.4=20\end{matrix}\right.\)

Vậy x = 25; y = 20.

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

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{7}{7}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=1\\\dfrac{y}{-5}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-5\end{matrix}\right.\)

Vậy x = 2; y = -5.

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

\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{19}=2\\\dfrac{y}{21}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2.19=38\\y=2.21=42\end{matrix}\right.\)

Vậy x = 38; y = 42.

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

Dựa vào tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{5}=\dfrac{y}{-2}=\dfrac{x+y}{5+-2}=\dfrac{30}{3}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.-2=-20\end{matrix}\right.\)

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

Dựa vào tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x+y}{5+3}=\dfrac{40}{8}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=5.3=15\end{matrix}\right.\)

\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}\Rightarrow\dfrac{3x}{15}=\dfrac{2y}{8}\)

Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x}{15}=\dfrac{2y}{8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=5.4=20\end{matrix}\right.\)

\(x:2=y:\left(-5\right)\Rightarrow\dfrac{x}{2}=\dfrac{y}{-5}\)

Dựa vào tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{7}{7}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.2=2\\y=1.\left(-5\right)=-5\end{matrix}\right.\)

\(\dfrac{x}{19}=\dfrac{y}{21}\Rightarrow\dfrac{2x}{38}=\dfrac{y}{21}\)

Dựa vào tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.19=38\\y=2.21=42\end{matrix}\right.\)

1,Ta có:-2x=5y\(\Rightarrow\)\(\dfrac{x}{5}\)=\(\dfrac{y}{-2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{x}{5}\)=\(\dfrac{y}{-2}\)=\(\dfrac{x+y}{5+-2}\)=\(\dfrac{30}{3}\)=10

=>x=-2.10=-20

=>y=5.10=50

Vậy x=-20;y=50

2,Ta có:3x=5y\(\Rightarrow\)\(\dfrac{x}{5}\)=\(\dfrac{y}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{x}{5}\)=\(\dfrac{y}{3}\)=\(\dfrac{x+y}{5+3}\)=\(\dfrac{40}{8}\)=5

=>x=3.5=15

=>y=5.5=25

Vậy x=15;y=25

Ta có :

\(ad=bc\left(1\right)\)

Chia cả 2 vế của \(\left(1\right)\) cho \(bd\) ta được :

\(VT=\dfrac{ad}{bd}=\dfrac{a}{b}\left(2\right)\)

\(VP=\dfrac{bc}{bd}=\dfrac{c}{d}\left(3\right)\)

Từ \(\left(2\right)+\left(3\right)\Leftrightarrowđpcm\)

Từ có đẳng thức: \(ad=bc\)

\(\Rightarrow\dfrac{ad}{cd}=\dfrac{bc}{cd}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\) (đpcm)

Giải:

\(\left(x^4\right)^3=\dfrac{x^{18}}{x^7}\)

\(\Leftrightarrow x^{12}=x^{11}\)

\(\Leftrightarrow x^{12}-x^{11}=0\)

\(\Leftrightarrow x^{11}\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^{11}=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy ...

\(\left(x^4\right)^3=\dfrac{x^{18}}{x^7}\Rightarrow x^{12}=x^{11}\)

<=> x¹²-x¹¹=0

<=> x¹¹(x-1)=0

=> \(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

vậy x=1

Có: \(\left(x^4\right)^3=\dfrac{x^{18}}{x^7}\)

⇒ \(x^{12}=x^9\)

⇒ \(x\in\left\{0;1\right\}\)

Vậy x = 0 hoặc 1.

Giải:

\(\dfrac{3^7.8^5}{6^6.\left(-2\right)^{12}}\)

\(=\dfrac{3^7.8^5}{6^6.2^{12}}\)

\(=\dfrac{3^7.2^{15}}{3^6.2^6.2^{12}}\)

\(=\dfrac{3^7.2^{15}}{3^6.2^{18}}\)

\(=\dfrac{3}{2^3}\)

\(=\dfrac{3}{8}\)

Vậy ...

\(\dfrac{3^7.8^5}{6^6.\left(-2\right)^{12}}=\dfrac{3^7.2^{15}}{2^6.3^6.2^{12}}=\dfrac{3}{8}\)

\(\dfrac{3^7.8^5}{6^6.\left(-2\right)^{12}}=\dfrac{3^7.\left(2^3\right)^5}{\left(2.3\right)^6.2^{12}}=\dfrac{3^7.2^{15}}{2^6.3^6.2^{12}}=\dfrac{3}{2^3}=\dfrac{3}{8}\)

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=t\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)

Khi đó:

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bt+b}{dt+d}\right)^2==\left[\dfrac{b\left(t+1\right)}{d\left(t+1\right)}\right]^2=\dfrac{b^2}{d^2}\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2t^2+b^2}{d^2t^2+d^2}=\dfrac{b^2\left(t^2+1\right)}{d^2\left(t^2+1\right)}=\dfrac{b^2}{d^2}\)

Ta có điều phải chứng minh

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

nên \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\left(\dfrac{a+b}{c+d}\right)^2\)

suy ra \(\dfrac{a^2+b^2}{c^2+d^2}=\left(\dfrac{a+b}{c+d}\right)^2\Rightarrow dpcm\)

a)đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k\(\Rightarrow\)a=bk, c=dk
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
từ (1),(2)\(\Rightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)

b)ta có:
\(\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
câu c bn tự giải nhé dễ mak ahihihichúc bn hc tốt