Từ giả thiết ta có \(\frac{15a-10b}{25}=\frac{6c-15}{9}=\frac{10b-6c}{4}\)

\(=\frac{0}{38}=0\)

(Theo t/c day ti so bang nhau)

Suy ra \(\hept{\begin{cases}15a-10b=0\\6c-15a=0\end{cases}\Rightarrow\hept{\begin{cases}3a-2b=0\\2c-5a=0\end{cases}}}\Rightarrow\hept{\begin{cases}b=\frac{3}{2}a\\c=\frac{5}{2}a\end{cases}}\)

Mà a^2+275=bc Suy ra \(^{a^2+275=\frac{15}{4}a^2\Rightarrow a^2=100\Rightarrow a=\pm10}\)

ĐS: a=10; b=15; c=25 và a=-10; b=-15; c=-25

Sửa chút \(\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}\)

Cho mik hỏi sao a mũ 2 2+275=15/4*a mũ 2 thì sao a mũ 2 lại bằng 100 ạ

a) Ta có:

+) a/2=b/3

=>a=2b/3

+) b/5=c/4

=>c=4b/5

Lại có:

a-b+c=49

=> 2b/3 -b + 4b/5 =49

=> 7b/15==49

=> b= 105

Khi đó:

+) a=2b/3=2.105/3=70

+)c=4b/5=4.105/5=84

Vậy a=70; b=105; c=84...

chúc bạn học tốt

b: 3a=2b nen a/2=b/3

5b=7c nên b/7=c/5

=>a/14=b/21=c/15

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

\(\dfrac{a}{14}=\dfrac{b}{21}=\dfrac{c}{15}=\dfrac{3a+5b-7c}{3\cdot14+5\cdot21-7\cdot15}=\dfrac{60}{42}=\dfrac{10}{7}\)

=>a=20; b=30; c=150/7

c: a/b=8/5 nên a/8=b/5

b/3=c/5 nên b/15=c/25

=>a/24=b/15=c/25=k

=>a=24k; b=15k; c=25k

a+b+c=61

=>24k+15k+25k=61

=>k=61/64

=>a=183/8; b=915/64; c=1525/64

e: a/2=b/3 nên a/8=b/12

b/4=c/5 nên b/12=c/15

=>a/8=b/12=c/15=k

=>a=8k; b=12k; c=15k

a^2-b^2=-16

=>64k^2-144k^2=-16

=>80k^2=16

=>k^2=1/5

TH1: k=1/căn 5

=>\(a=\dfrac{8}{\sqrt{5}};b=\dfrac{12}{\sqrt{5}};c=3\sqrt{5}\)

TH2: k=-1/căn 5

=>\(a=\dfrac{-8}{\sqrt{5}};b=\dfrac{-12}{\sqrt{5}};c=-3\sqrt{5}\)

Ta có:

\(Q=\dfrac{2a-b}{3a-b}+\dfrac{5b-a}{3a+b}\)

\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)}{\left(3a-b\right)\left(3a+b\right)}+\dfrac{\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)

\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)

\(Q=\dfrac{3a^2+15ab-6b^2}{9a^2-b^2}\)

Ta lại có:

\(6a^2-15ab+5b^2=0\)

\(\Rightarrow3a^2+15ab-6b^2=9a^2-b^2\left(1\right)\)

Thay (1) vào Q

=> Q = 1

Lời giải:

\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)

\(\Leftrightarrow \frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}\)

Áp dụng TC dãy tỉ số bằng nhau:

\(\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}=\frac{15a-10b+6c-15a+10b-6c}{25+9+4}=0\)

\(\Rightarrow 15a=10b=6c\)

\(\Leftrightarrow \frac{a}{6}=\frac{b}{9}=\frac{c}{15}\)

Tiếp tục áp dụng TCDTSBN:

$\frac{a}{6}=\frac{b}{9}=\frac{c}{15}=\frac{a+b+c}{6+9+15}=\frac{50}{30}=\frac{5}{3}$

$\Rightarrow a=10; b=15; c=25$

bạn tham khảo tại: câu hỏi của nguyễn thị thanh mai- onlinemath

Biểu thức có giá trị bằng 2 thì:

a) \(\dfrac{3a^2}{10b^3}\cdot\dfrac{15b}{9a^4}\)

\(=\dfrac{3a^2\cdot15b}{10b^3\cdot9a^4}\)

\(=\dfrac{1\cdot3}{2\cdot b^2\cdot3\cdot a^2}=\dfrac{3}{6a^2b^2}\)

b) \(\dfrac{x-3}{x^2}\cdot\dfrac{4x}{x^2-9}\)

\(=\dfrac{x-3}{x^2}\cdot\dfrac{4x}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{\left(x-3\right)\cdot4x}{x^2\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{4}{x\left(x+3\right)}\)

c) \(\dfrac{a^2-6x+9}{a^2+3a}\cdot\dfrac{2a+6}{a-3}\)

\(=\dfrac{\left(a-3\right)^2}{a\left(a+3\right)}\cdot\dfrac{2\cdot\left(a+3\right)}{a-3}\)

\(=\dfrac{\left(a-3\right)^2\cdot2\cdot\left(a+3\right)}{a\left(a+3\right)\left(a-3\right)}\)

\(=\dfrac{2\left(a-3\right)}{a}\)

d) \(\dfrac{x+1}{x}\cdot\left(x+\dfrac{2-x^2}{x^2-1}\right)\)

\(=\dfrac{\left(x+1\right)\cdot x}{x}+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{x^2-1}\)

\(=x+1+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{\left(x+1\right)\left(x-1\right)}\)

\(=x+\dfrac{2-x^2}{x\left(x-1\right)}\)