\(\dfrac{2}{3}\times\dfrac{3}{4}+\dfrac{3}{4}\times\dfrac{2}{5}=\dfrac{3}{4}\times\left(\dfrac{2}{3}+\dfrac{2}{5}\right)=\dfrac{3}{4}\times\dfrac{16}{15}=\dfrac{4}{5}\)

0.8

0,8

Bài 1: 

=>y-69,3=20,5

hay y=89,8

Bài 2: 

a: \(=365.4:43.5-7.7\cdot0.2\)

=8,4-1,54=6,86

b: \(=\dfrac{2}{5}:\dfrac{8-5}{10}+\dfrac{3}{5}=\dfrac{2}{5}\cdot\dfrac{10}{3}+\dfrac{3}{5}=\dfrac{20+9}{15}=\dfrac{29}{15}\)

Giải:

a) \(11\dfrac{3}{4}.\left(6\dfrac{5}{6}-4\dfrac{1}{2}+1\dfrac{2}{3}\right)\) 

\(=\dfrac{47}{4}.\left(\dfrac{41}{6}-\dfrac{9}{2}+\dfrac{5}{3}\right)\) 

\(=\dfrac{47}{4}.4\) 

\(=47\) 

b) \(\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\) 

\(=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\) 

\(=\dfrac{25}{8}:\dfrac{75}{26}\) 

\(=\dfrac{13}{12}\)

c) \(\left(17\dfrac{13}{15}-3\dfrac{3}{7}\right)-\left(2\dfrac{12}{15}-4\right)\) 

\(=\dfrac{268}{15}-\dfrac{24}{7}-\dfrac{14}{5}+4\) 

\(=\left(\dfrac{268}{15}-\dfrac{14}{5}\right)+\left(\dfrac{-24}{7}+4\right)\) 

\(=\dfrac{226}{15}+\dfrac{4}{7}\) 

\(=\dfrac{1642}{105}\) 

d) \(2\dfrac{2}{3}.\left(\dfrac{-4}{5}.0,375.-10.\dfrac{-15}{24}\right)\) 

\(=\dfrac{8}{3}.\left(\dfrac{-4}{5}.\dfrac{3}{8}.-10.\dfrac{-5}{8}\right)\) 

\(=\left(\dfrac{8}{3}.\dfrac{3}{8}\right).\left(\dfrac{-4}{5}.\dfrac{-5}{8}.-10\right)\) 

\(=1.-5\) 

\(=-5\) 

Chúc bạn học tốt!

a, = 8/15 x 5/6 = 40/90 = 4/9

b, = 5/6 x 1/5 = 5/30 = 1/6

a, \(= 8/15 x 5/6 = 40/90 = 4/9\)

b, \(= 5/6 x 1/5 = 5/30 = 1/6\)

G(-2) = (-2)² – 4 = 4 – 4 = 0;

G(1) = 1² – 4 = 1 – 4 = -3;

G(0) = 0² – 4 = 0 – 4 = -4;

G(1) = 1² – 4 = 1- 4 = -3;

G(2) = 2² – 4 = 4 – 4 = 0

Lời giải:

$a=2+\sqrt{5}$

$a-2=\sqrt{5}$

$a^2-4a+4=5\Leftrightarrow a^2-4a-1=0$

$p(a)=a^5-13a^4+7a^3-4a^2-6a$

$=a^3(a^2-4a-1)-9a^2(a^2-4a-1)-28a(a^2-4a-1)-125a^2-34a$

$=-125a^2-34a=-125(a^2-4a-1)-534a-125$

$=-534a-125=-534(2+\sqrt{5})-125=-1193-534\sqrt{5}$

1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)

\(\Leftrightarrow2x^2+6x-6x+18=0\)

\(\Leftrightarrow2x^2+18=0\left(loại\right)\)

2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)

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

4: Ta có: \(2x\left(x-5\right)-3x+15=0\)

\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)

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

5: Ta có: \(3x\left(x+4\right)-2x-8=0\)

\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)

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

6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3

1)  (2x + 1)(3x – 2) = (5x – 8)(2x + 1)

⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0

⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0

⇔ (2x + 1).(3x – 2 – 5x + 8) = 0

⇔ (2x + 1)(6 – 2x) = 0

⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)

Vậy.....

2)  4x² -1 = (2x + 1)(3x - 5)

⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0

⇔ (2x+1)(2x-1-3x+5)=0

⇔ (2x+1)(4-x)=0

⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)

Vậy...

3)  

(x + 1)² = 4(x² – 2x + 1)

⇔ (x + 1)² - 4(x² – 2x + 1) = 0

⇔ x² + 2x +1- 4x² + 8x – 4 = 0

⇔ - 3x² + 10x – 3 = 0

⇔ (- 3x² + 9x) + (x – 3) = 0

⇔ -3x (x – 3)+ ( x- 3) = 0

⇔ ( x- 3) ( - 3x + 1) = 0

⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy......

4) 2x³+5x²-3x=0

⇒2x³-x²+6x²-3x=0

⇒(2x³-x²)+(6x²-3x)=0

⇒x²(2x-1)+3x(2x-1)=0

⇒(x²+3x)(2x-1)=0

⇒ hoặc x²+3x=0⇒x(x+3)=0⇒hoặc x=0 hoặc x=-3

hoặc 2x-1=0⇒x=0,5

Vậy ...

5)2x=3x-2

⇒2x-3x=-2

⇒-x=-2

⇒x=2

6) x+15=3x-1

⇒x-3x=-1-15

⇒-2x=-16

⇒x=8

7)2-x=0,5x-4

⇒-x-0,5x=-4-2

⇒-1,5x=-6

⇒x=4