a.\(\left(3x-2\right)^2=16\)

Ta có: \(\left(3x-2\right)^2=16\)

\(\Rightarrow\left(3x-2\right)^2=\left(4\right)^2\)

\(\Rightarrow3x-2=4\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

b. \(\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\dfrac{-8}{125}\)

\(\Rightarrow\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\left(\dfrac{-2}{5}\right)^3\)

\(\Rightarrow\dfrac{4}{5}x-\dfrac{3}{4}=\dfrac{-2}{5}^{ }\)

\(\Rightarrow\dfrac{4}{5}x-=\dfrac{7}{20}\)

\(\Rightarrow x=\dfrac{7}{16}\)

c. \(5^{x+2}+5^x=3250\)

\(5^x.5^2+5^x=3250\)

\(5^x.\left(5^2+1\right)=3250\)

\(5^x.26=3250\)

\(5^x=125\)

\(5^x=5^3\)

\(x=5\)

\(\dfrac{5}{6}x-\dfrac{3}{4}=\dfrac{-1}{4}+\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{5}{6}x=\dfrac{7}{6}\)

\(\Rightarrow x=\dfrac{7}{5}\)

b) \(-1\dfrac{1}{2}-\dfrac{2}{3}x=\dfrac{5}{6}-\left(\dfrac{-2}{5}\right)\)

\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{41}{15}\)

\(\Rightarrow x=-\dfrac{41}{10}\)

c) \(\left(\dfrac{4}{5}:x+1,5\right):\dfrac{2}{3}=-1,5\)

\(\Leftrightarrow\dfrac{8+15x}{10x}.\dfrac{3}{2}=\dfrac{-3}{2}\)

\(\Leftrightarrow\dfrac{24+45x}{20x}=\dfrac{-3}{2}\)

\(\Leftrightarrow-60x=48+90x\)

\(\Rightarrow x=-0,32\)

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

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

\(\Rightarrow16x-8=3-12x\)

\(\Rightarrow x=\dfrac{11}{28}\)

a) Vì

)

A=\(\dfrac{9}{4}-\left(\dfrac{-5}{2}\right)+\left(\dfrac{-25}{6}\right)\)

A=\(\dfrac{19}{4}\)-\(\dfrac{25}{6}\)

A=\(\dfrac{14}{24}\)=\(\dfrac{7}{12}\)

b, Thay vào, ta sẽ có:

A=3.\(\left(\dfrac{-2}{3}\right)-4.\left(\dfrac{-2}{3}\right).\dfrac{4}{5}+5.\dfrac{4}{5}\)

A=-2-\(\left(\dfrac{-32}{15}\right)\)+4

A=\(\dfrac{2}{15}\)+4

A=\(\dfrac{62}{15}\)

a: \(\dfrac{2.75}{x}=\dfrac{0.4}{1.5}=\dfrac{4}{15}\)

\(\Leftrightarrow x=\dfrac{11}{4}\cdot\dfrac{15}{4}=\dfrac{165}{16}\)

b: \(3\dfrac{1}{2}:\left(2x-3\right)=\dfrac{-3}{4}:0.2\)

\(\Leftrightarrow\dfrac{7}{2}:\left(2x-3\right)=\dfrac{-3}{4}:\dfrac{1}{5}=\dfrac{-15}{4}\)

\(\Leftrightarrow2x-3=\dfrac{7}{2}:\dfrac{-15}{4}=\dfrac{-7}{2}\cdot\dfrac{4}{15}=\dfrac{-28}{30}=\dfrac{-14}{15}\)

=>2x=-14/15+3=45/45-14/15=31/45

=>x=31/90

c: \(\dfrac{3x+2}{27}=\dfrac{3}{3x+2}\)

\(\Leftrightarrow\left(3x+2\right)^2=81\)

=>3x+2=9 hoặc 3x+2=-9

=>3x=7 hoặc 3x=-11

=>x=7/3 hoặc x=-11/3

d: \(\dfrac{5-x}{4}=\dfrac{2x+3}{2}\)

=>10-2x=8x+12

=>-10x=2

hay x=-1/5

c) +)Điểm A ( 1;9) => x = 1 ; y = 9

Thay x = 1 vào y = 4x+5 , ta có:

y = 4.1+5

y = 4+5

y = 9

Vậy điểm A ( 1;9 ) thuộc đồ thị hàm số y = 4x +5

+) Điểm B ( -2;3 ) => x = -2 ; y = 3

Thay x = -2 vào y = 4x +5 , ta có:

y = 4.(-2) + 5

y = (-8) + 5

y = (-3)

Vậy điểm B ( -2;3) không thuộc đồ thị hàm số y = 4x+5

....Các câu khác tương tự....> . <...

1 )Ta có

\(M=\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right).....\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{4}{3}\cdot\dfrac{-3}{4}\cdot\dfrac{5}{4}\cdot\cdot\cdot\cdot\dfrac{-99}{100}\cdot\dfrac{101}{100}\)

\(=\dfrac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot3\cdot\left(-4\right)\cdot4\cdot\left(-5\right)\cdot5....\cdot\left(-100\right)\cdot100\cdot101}{2^2\cdot3^2\cdot4^2....\cdot100^2}\)

\(=-\dfrac{101}{200}< \dfrac{1}{2}\)

2 ) Số phân số của biểu thức B là 180 phân số

Ta có

\(\dfrac{1}{20}>\dfrac{1}{200};\dfrac{1}{21}>\dfrac{1}{200};\dfrac{1}{22}>\dfrac{1}{200};....;\dfrac{1}{199}>\dfrac{1}{200}\)

\(\Rightarrow B=\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}\cdot180=\dfrac{9}{10}\)

\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)

=1/64

a. Ta có : ( a + b )( c - d ) = ac-ad+bc-bd (1)

( a - b )( c + d ) = ac+ad-bc+bd (2)

Từ giả thuyết : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\) (3)

Từ (1) , ( 2) và ( 3) \(\Rightarrow\)( a + b )( c - d) = ( a - b)( c + d )

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)