Bài 2:

\(\Leftrightarrow\left\{{}\begin{matrix}3^{3x-2x+y}=3^5\\5^{2x-x-y}=5^3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\x-y=3\end{matrix}\right.\)

=>x=1;y=-2

a) => 5^x.5² + 5^x.5³=750

=> 5^x . (5²+5³) =750

=> 5^x . 150 =750

=> 5^x = 750 : 150

=> 5^x = 5

=> x =1

Vậy x = 1

b) => 3^2x+1 . 7^y = 3² . (3.7)^x

=> 3^2x+1 . 7^y = 3^x+2 . 7^x

=> \(\dfrac{3^{2x+1}}{3^{x+2}}\) =\(\dfrac{7^x}{7^y}\)

=> 3^(2x+1)-(x+2) = 7^x-y

=> 3^x-1 = 7^x-y

=>^{\(\left\{{}\begin{matrix}x-1=0\\x-y=0\end{matrix}\right.\)}

=>\(\left\{{}\begin{matrix}x=1\\x=y\end{matrix}\right.\)

=>x=y=1

Vậy x=y=1

c)

=>\(\dfrac{3^{3x}}{3^{2x-y}}\) =3^{5 và} =>\(\dfrac{5^{2x}}{5^{x+y}}\) =5³

=> 3(3x)-(2x-y) =3⁵ =>5^(2x)-(x+y) =5³

=> 33x-2x+y =3⁵ => 5^2x-x-y =5³

=> 3x+y =3⁵ => 5^x-y =5³

=> x+y =5 (1) => x-y =3 (2)

Từ (1) và (2) có :

+x = (5+3):2 =4

+y = (5-3):2 =1

Vậy x=4 ; y=1

- Nếu làm đúng cho mình xin cái tick ! Tks

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

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

⇒\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)⇒\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)

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

\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)

⇒x=42,y=28,z=20

\(\dfrac{x}{3}=\dfrac{y}{2}\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}\)

\(\dfrac{x}{5}=\dfrac{z}{7}\text{⇒}\dfrac{x}{15}=\dfrac{z}{21}\)

⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{21}\)⇒\(\dfrac{x}{15}=\dfrac{2y}{20}\)

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

\(\dfrac{x}{15}=\dfrac{2y}{20}=\dfrac{x+2y}{15+20}=\dfrac{-112}{35}=\dfrac{-16}{5}\)

⇒x=48,y=32,z=336/5

Lời giải:

1. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{3}=\frac{y}{2}=\frac{2y}{4}=\frac{x+2y}{3+4}=\frac{-112}{7}=-16$

$\Rightarrow x=-16.3=-48; y=-16.2=-32$

Đoạn $\frac{x}{5}=\frac{x}{7}$ là sao em? Em xem lại đề.

2. 

$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}\Rightarrow \frac{x}{21}=\frac{y}{14}(1)$

$5y=7z\Rightarrow \frac{y}{7}=\frac{z}{5}\Rightarrow \frac{y}{14}=\frac{z}{10}(2)$

Từ $(1);(2)\Rightarrow \frac{x}{21}=\frac{y}{14}=\frac{z}{10}$

Áp dụng tính chất dãy tỷ số bằng nhau:

$\frac{x}{21}=\frac{y}{14}=\frac{z}{10}$

$=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2$

$\Rightarrow x=2.21=42; y=2.14=28; z=2.10=20$

1)

a) \(0,25^x\cdot12^x=243\)

\(\Leftrightarrow\left(0,25\cdot12\right)^x=3^5\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

b) \(38^y:19^y=512\)

\(\Leftrightarrow2y\cdot y=512\)

\(\Leftrightarrow2y^2=512\)

\(\Leftrightarrow y^2=256\)

\(\Leftrightarrow\left[{}\begin{matrix}y=16\\y=-16\end{matrix}\right.\)

Vậy \(y_1=-16;y_2=16\)

2)

a) \(3^x+3^{x+2}=2430\)

\(\Leftrightarrow\left(1+3^2\right)\cdot3^x=2430\)

\(\Leftrightarrow\left(1+9\right)\cdot3^x=2430\)

\(\Leftrightarrow10\cdot3^x=2430\)

\(\Leftrightarrow3^x=243\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

b) \(2^{x+3}-2^x=224\)

\(\Leftrightarrow\left(2^3-1\right)\cdot2^x=224\)

\(\Leftrightarrow\left(8-1\right)\cdot2^x=224\)

\(\Leftrightarrow7\cdot2^x=224\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

3)

a) \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow x-\dfrac{1}{4}=\pm\dfrac{2}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{2}{3}\\x-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}+\dfrac{1}{4}\\x=-\dfrac{2}{3}+\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=-\dfrac{5}{12}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{11}{12};x_2=-\dfrac{5}{12}\)

b) \(\left(x+0,7\right)^3=-27\)

\(\Leftrightarrow\left(x+\dfrac{3}{10}\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x+\dfrac{3}{10}=-3\)

\(\Leftrightarrow x=-3-\dfrac{3}{10}\)

\(\Leftrightarrow x=-\dfrac{37}{10}\)

Vậy \(x=-\dfrac{37}{10}\)

4)

a) \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\dfrac{2}{5}-3x=\pm\dfrac{3}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{1}{15};x_2=\dfrac{1}{3}\)

b) \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)

\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\)

\(\Leftrightarrow2x-1=1\)

\(\Leftrightarrow2x=1+1\)

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

1. a) \(0,25^x.12^x=243\)

\(\Rightarrow\left(0,25.12\right)^x=243\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

b) \(38^y:19^y=512\)

\(\Rightarrow\left(38:19\right)^y=512\)

\(\Rightarrow2^y=2^9\)

\(\Rightarrow y=9\)

Vậy \(y=9.\)

2) a) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x\left(1+9\right)=2430\)

\(\Rightarrow3^x=243=3^5\)

\(\Rightarrow x=5\)

Vậy x=5.

b) \(2^{x+3}-2^x=224\)

\(\Rightarrow2^x\left(8-1\right)=224\)

\(\Rightarrow2^x=32=2^5\)

\(\Rightarrow x=5\)

Vậy x=5.

Bài 3: dễ tự làm.