Ta có: \(\frac{3^{31}.25^8-9^{14}.5^{15}}{27^{10}.5^{15}-3^{30}.5^{17}}=\frac{3^{31}.\left(5^2\right)^8-\left(3^2\right)^{14}.5^{15}}{\left(3^3\right)^{10}.5^{15}-3^{30}.5^{17}}=\frac{3^{31}.5^{16}-3^{28}.5^{15}}{3^{30}.5^{15}-3^{50}.5^{17}}\)

\(=\frac{5^{28}.5^{15}.\left(3^3.5-1\right)}{3^{30}.5^{15}.\left(1-5^2\right)}=\frac{135-1}{3.\left(-24\right)}=\frac{134}{-72}=-\frac{67}{36}\)

Ta có: \(\left(\frac{1}{16}\right)^x=\left(\frac{1}{2}\right)^{20}\) 

=> \(\left[\left(\frac{1}{2}\right)^4\right]^x=\left(\frac{1}{2}\right)^{20}\)

=> \(\left(\frac{1}{2}\right)^{4x}=\left(\frac{1}{2}\right)^{20}\)

=> 4x = 20

=> x = 20 : 4

=> x=  5

\(P=0,4\left(3\right)+0,6\left(2\right)\cdot2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:\frac{50}{53}\)

\(\Rightarrow P=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{3+2}{6}:\frac{53}{90}\cdot\frac{53}{50}\)

\(\Rightarrow P=\frac{13}{30}+\frac{14}{9}-\frac{5}{6}\cdot\frac{90}{53}\cdot\frac{53}{50}\)

\(\Rightarrow P=\frac{39}{90}+\frac{140}{90}-\frac{2}{3}\)

\(\Rightarrow\frac{179}{90}-\frac{60}{90}=\frac{119}{90}\)

\(\frac{\left[\left(5,2^2:2,6+8,1\right)^2-6,5^2\right]:0,025}{\left(60,192:2,4-1,08\right)^2-0,24.140}\)

\(=\frac{\left[\left(27,04:2,6+8,1\right)^2-42,25\right]:0,025}{\left(25,08-1,08\right)^2-336}\)

\(=\frac{\left[\left(10,4+8,1\right)^2-42,25\right]:0,025}{24^2-336}\)

\(=\frac{\left(18,5^2-42,25\right):0,025}{576-336}\)

\(=\frac{\left(342,25-42,25\right):0,025}{240}\)

\(=\frac{300:0,025}{240}=\frac{12000}{240}=50\)

a.

\(\hept{\begin{cases}x=\frac{-23}{68}=\frac{-23.172}{68.172}=\frac{-3956}{11696}\\y=\frac{-57}{172}=\frac{-57.58}{172.68}=\frac{-3306}{11696}\end{cases}\Leftrightarrow x>y}\)

b.

\(\hept{\begin{cases}x=\frac{-31}{33}=\frac{-31.51}{33.51}=\frac{-1581}{1683}\\y=\frac{-49.33}{51.33}=\frac{-1617}{1683}\end{cases}}\Leftrightarrow x>y\)

Sửa câu a: \(x< y\)

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

\(\frac{x}{6}=\frac{y}{9}=\frac{x-y}{6-9}=\frac{30}{-3}=-10\)

\(\Rightarrow x=-60;y=-90\)

a, Ta có: \(\frac{a}{c}\)= \(\frac{c}{b}\)\(\Rightarrow\)\(ab\)= \(c^2\)

Để chứng minh \(\frac{a^2+c^2}{b^2+c^2}\)= \(\frac{a}{b}\)thì ta phải chứng minh b(a²+c²)=a(b²+c²)

Ta có: b(a²+c²)= b.a²+b.c² (1)

Thay ab= c² vào 1 ta có:

b.a²+b.a.b= b².a+a².bb

Ta có: a(b²+c²) = a.b²+a.c² (2)

Thay ab= c² vào (1) ta có:

a.b²+b.a.a= b².a+a².bb

Vì b².a+a².b= b².a+a².b \(\Rightarrow\)b(a²+c²)= a(b²+c²)

\(\Rightarrow\)\(\frac{a^2+c^2}{b^2+c^2}\)= \(\frac{a}{b}\)

\(\Rightarrow\)Đpcm (Điều phải chứng minh)

Chúc bn học tốt

a.

\(\frac{a}{c}=\frac{c}{b}\Leftrightarrow c^2=ab\Rightarrow\frac{a^2+ab}{b^2+ab}=\frac{a.\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\)

b.

\(\frac{a}{c}=\frac{c}{b}\Leftrightarrow c^2=ab\Rightarrow\frac{\left(b^2-ab\right)+\left(ab-a^2\right)}{a\left(a+b\right)}=\frac{b\left(b-a\right)+a\left(b-a\right)}{a\left(a+b\right)}=\frac{b-a}{a}\)

Ta có : c/5=2c/10

Lại có a-b+2c=77

Áp dụng tính chất của dãy các tỉ số bằng nhau

Ta có a/3=b/2=2c/10=a-b+2c/3-2+10=77/11=7

=>a/3=7=>a=21

b/2=7=>b=14

2c/10=7=>c=10.7:2=35

Vậy a,b,c lần lượt là 21,14,35

Học tốt

T cho mk nhé