14 phút 25 giây x 4= 57 phút

4 phút 25 giây × 4 = 17 phút 40 giây 

a) x vô nghĩa

b) x=0,8;x=-0,8

d: Đặt x/2=y/5=k

=>x=2k; y=5k

Ta có: xy=10

nên k²=1

Trường hợp 1: k=1

=>x=2; y=5

Trường hợp 2: k=-1

=>x=-2; y=-5

( - 25 ) : 10 = -15 : x

=> -15 : x = - 25 : 10

     -15 : x = - 2,5

     x          = -15 : -2,5

     x          = 6

Vậy x = 6

Ta có: (-25) : 10 = (-15) : x

=>      -2,5          = (-15) : x

=>       x             = (-15) : (-2,5)

=>       x             = 6

Vậy x = 6 

-25 : 10 = -15 : x

=> -25x = -15 . 10

-25x = -150

x= 6

a)x*5,2:4,5=10,4

x*5,2=10,4*4,5

x*5,2=46,8

x      =46,8:5,2

x      =9

b)5/7*x+2/7*x=2/3

x*(5/7+2/7)=2/3

x*1            =2/3

x                =2/3:1

x                 =2/3

a. 

\(x\times5,2\div4,5=10,4\)

\(x\times5,2=10,4\times4,5\)

\(x\times5,2=46,8\)

\(x=46,8\div5,2\)

\(x=9\)

b.

\(\dfrac{5}{7}\times x+\dfrac{2}{7}\times x=\dfrac{2}{3}\)

\(x\times\left(\dfrac{5}{7}+\dfrac{2}{7}\right)=\dfrac{2}{3}\)

\(x\times1=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}\div1\)

\(x=\dfrac{2}{3}\)

vào câu hỏi tương tự khác có

 Ta có:\(VT=\left(x+a\right)\left(x+5\right)\)

\(=x^2+\left(a+5\right)x+5a\)

Và \(VP=x^2+3x+b\)

Đồng nhất 2 vế ta có: \(\hept{\begin{cases}a+5=3\\b=5a\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-2\\b=-10\end{cases}}\)

\(B=\frac{4^2.25^2+32.125}{2^3.5^2}\)

\(=\frac{\left(2^2\right)^2.\left(5^2\right)^2+2^5.5^3}{2^3.5^2}\)

\(=\frac{2^4.5^4+2^5.5^3}{2^3.5^2}\)

\(=\frac{2^3.5^2.\left(2.5^2+2^2.5\right)}{2^3.5^2}\)

\(=2.5^2+2^2.5\)

\(=2.25+4.5\)

\(=50+20\)

\(=70\)

Bài làm

\(B=\frac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\)

\(B=\frac{\left(2^2\right)^2\cdot\left(5^2\right)^2+2^5\cdot5^3}{2^3\cdot5^2}\)

\(B=\frac{2^4\cdot5^2+2^5\cdot5^3}{2^3\cdot5^2}\)

\(B=\frac{2^4\left(5^2+2\cdot5^3\right)}{2^3.5^2}\)

\(B=\frac{2^4\left[5^2\left(1+2\cdot5\right)\right]}{2^3.5^2}\)

\(B=\frac{2^4\cdot5^2\cdot11}{2^3\cdot5^2}\)

\(B=2.11=22\)

Vậy B = 22

Xl, mik làm bị lỗi rồi nha. 

`a)`

`A=(x+1)(2x-1)`

`=2x^{2}+x-1`

`=2(x^{2}+(1)/(2)x-(1)/(2))`

`=2(x^{2}+(1)/(2)x+(1)/(16)-(9)/(16))`

`=2(x+(1)/(4))^{2}-(9)/(8)>= -9/8` với mọi `x`

Dấu `=` xảy ra khi :

`x+(1)/(4)=0<=>x=-1/4`

Vậy `min=-9/8<=>x=-1/4`

``

`b)`

`(4x+1)(2x-5)`

`=8x^{2}-18x-5`

`=8(x^{2}-(9)/(4)x-(5)/(8))`

`=8(x^{2}-(9)/(4)x+(81)/(64)-(121)/(64))`

`=8(x-(9)/(8))^{2}-(121)/(8)>= -(121)/(8)` với mọi `x`

Dấu `=` xảy ra khi :

`x-(9)/(8)=0<=>x=9/8`

Vậy `min=-121/8<=>x=9/8`

\(A=2x^2+x-1=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

\(A_{min}=-\dfrac{9}{8}\) khi \(x=-\dfrac{1}{4}\)

\(B=8x^2-18x-5=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)

\(B_{min}=-\dfrac{121}{8}\) khi \(x=\dfrac{9}{8}\)

a) \(A=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{9}{8}=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\)Vì \(2\left(x+\dfrac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

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

b) \(B=\left(4x+1\right)\left(2x-5\right)=8x^2-18x-5=8\left(x^2-\dfrac{9}{4}x+\dfrac{81}{64}\right)-\dfrac{121}{8}=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\)

Vì \(8\left(x-\dfrac{9}{8}\right)^2\ge0\Rightarrow8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)

\(ĐTXR\Leftrightarrow x=\dfrac{9}{8}\)