a) (x - 3)^x - (x - 3)^{x + 2} = 0

(x - 3)^x - (x - 3)^x . (x - 3)² = 0

(x - 3)^x.(1 - (x - 3)²) = 0

=> (x - 3)^x = 0     hoặc    1 - (x - 3)^x = 0

=> x - 3 = 0         hoặc    (x - 3)^x = 1

=> x = 3   

Thay x = 3 ở trường hợp 1 vào trường hợp 2

=. x - 3 = 1

=> x = 4

Câu 1:

\(3\left(x-1\right)=2\left(y-2\right)\Leftrightarrow3x-3=2y-4\Leftrightarrow3x=2y-1\)

\(4\left(y-2\right)=3\left(z-3\right)\Leftrightarrow4y-8=3z-9\Leftrightarrow4y=3z-1\)

Lại có:

\(3x=2y-1\Leftrightarrow6x=4y-2=3z-1-2=3z-3\)

\(\Rightarrow6x=4y-2=3z-3\)

\(\Rightarrow6x=3z-3\Leftrightarrow2x=z-1\)

\(\Rightarrow2x+3y-z=z-1+3y-z=3y-1=50\Leftrightarrow3y=51\Leftrightarrow y=17\)\(\Rightarrow\left\{{}\begin{matrix}x=11\\z=23\end{matrix}\right.\)

Câu 3:

\(\frac{a}{b}=\frac{8}{5}\Leftrightarrow\frac{a}{8}=\frac{b}{5}\Leftrightarrow\frac{1}{2}.\frac{a}{8}=\frac{1}{2}.\frac{b}{5}\Leftrightarrow\frac{a}{16}=\frac{b}{10}\) (1)

\(\frac{b}{c}=\frac{2}{7}\Leftrightarrow\frac{b}{2}=\frac{c}{7}\Leftrightarrow\frac{1}{5}.\frac{b}{2}=\frac{1}{5}.\frac{c}{7}\Leftrightarrow\frac{b}{10}=\frac{c}{35}\) (2)

Từ (1) và (2)

\(\Rightarrow\frac{a}{16}=\frac{b}{10}=\frac{c}{35}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=16k\\b=10k\\c=35k\end{matrix}\right.\)

\(\Rightarrow a+b+c=16k+10k+35k=61k=61\Rightarrow k=1\)

\(\Rightarrow\left\{{}\begin{matrix}a=16k=16\\b=10k=10\\c=35k=35\end{matrix}\right.\)

Câu 4:

\(ab=\frac{1}{2};bc=\frac{2}{3}\Leftrightarrow abbc=\frac{1}{2}.\frac{2}{3}\Leftrightarrow ab^2c=\frac{1}{3}\)

Lại có:

\(ac=\frac{3}{4}\Leftrightarrow b^2=\frac{4}{9}\Leftrightarrow\left[{}\begin{matrix}b=\frac{2}{3}\\b=-\frac{2}{3}\end{matrix}\right.\)

TH1: \(b=\frac{2}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}a=\frac{3}{4}\\c=1\end{matrix}\right.\)

TH2:\(b=-\frac{2}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}a=-\frac{3}{4}\\c=-1\end{matrix}\right.\)

dựa vào dạng toán dãy tỉ số bằng nhau

pạn trình bày cho mk tham khảo vs

Các bạn trình bày ra đi

3(x-2)-4(2x+1)-5(2x+3)=50

<=>(3x-6)-(8x+4)-(10x+15)=50

<=>3x-6-8x-4-10x-15=50

<=>(3x-8x-10x)+(-6-4-15)=50

<=>-15x-25=50

<=>-15x=75

<=>x=-5

\(3\frac{1}{2}:\left(4-\frac{1}{3}\left|2x+1\right|\right)=\frac{21}{22}\)

<=>\(4-\frac{1}{3}\left|2x+1\right|=\frac{7}{2}:\frac{21}{22}=\frac{11}{3}\)

<=>\(\frac{1}{3}\left|2x+1\right|=4-\frac{11}{3}=\frac{1}{3}\)

<=>\(\left|2x+1\right|=1\)

<=>2x+1=1 hoặc 2x+1=-1

<=>2x=0 hoặc 2x=-2

<=>x=0 hoặc x=-2

Vậy......................

\(\frac{3x-2y}{37}=\frac{5y-3z}{15}=\frac{2z-5x}{2}\)

\(\Leftrightarrow\frac{5\left(3x-2y\right)}{5.37}=\frac{2\left(5y-3z\right)}{2.15}=\frac{3\left(2z-5x\right)}{3.2}\)

\(\Leftrightarrow\frac{15x-10y}{5.37}=\frac{10y-6z}{2.15}=\frac{6z-15x}{3.2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\Rightarrow\frac{15x-10y}{5.37}=\frac{10y-6z}{2.15}=\frac{6z-15x}{3.2}=\frac{15x-10y+10y-6z+6z-15x}{5.37+2.15+3.2}=0\)

\(\Rightarrow\hept{\begin{cases}\frac{3x-2y}{37}=0\\\frac{5y-3z}{15}=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x-2y=0\\5y-3z=10\end{cases}\Leftrightarrow}\hept{\begin{cases}3x=2y\\5y=3z\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{5}\end{cases}\Leftrightarrow}\frac{x}{2}=\frac{y}{3}=\frac{z}{5}}\)

\(\Leftrightarrow\frac{10x}{20}=\frac{3y}{9}=\frac{2z}{10}\)

Áp dụng tính của dãy tỉ số bằng nhau:

\(\Leftrightarrow\frac{10x}{20}=\frac{3y}{9}=\frac{2z}{10}=\frac{10x-3y-2z}{20-9-10}=-\frac{4}{1}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=-4\\\frac{y}{3}=-4\\\frac{z}{5}=-4\end{cases}\Leftrightarrow\hept{\begin{cases}x=-8\\y=-12\\z=-20\end{cases}}}\)