|x|=-26:-13

|x|=2

x=-2      x=2

ủng hộ bạn tui nhá:)https://www.youtube.com/channel/UCb2H-q6FmW61PgcsL1OGPfw

\(-13.\left|x\right|=-26\)

\(\left|x\right|=\frac{-26}{-13}\)

\(\left|x\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

hok tốt!!

\(-13\left|x\right|=-26\)

\(\Leftrightarrow\left|x\right|=\left(-26\right):\left(-13\right)\)

\(\Leftrightarrow\left|x\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy\(x\in\left\{-2;2\right\}\)

Good girl

Tổng 2 số: 30*2=60

Số thứ 1: 60/(4+1)*4=48

Số thứ 2: 60-48=12

Vậy St1=48, St2=12

Tổng 2 số là :

   30 x 2 = 60

Ta có sơ đồ :

St1 :|^___|^___|^___|^___|              tổng : 60

St2 :|^___|

Tổng số phần bằng nhau là :

   4 + 1 = 5 (phần)

St1 là :

   60 : 5 x 4 = 48

St2 là :

   60 - 48 = 12 

       Đ/S : st1 : 48

                 st2 : 12

Cho ti le ban do 1:100.000,san truong co dien h. 36m2. Vay dt that cua san truong bao nhieu?

1) \(B=\dfrac{\sqrt{x}-5}{\sqrt{x}}\)

Thay \(x=\dfrac{4}{25}\) vào B, ta được:

\(B=\dfrac{\sqrt{\dfrac{4}{25}}-5}{\sqrt{\dfrac{4}{25}}}\)

\(=\dfrac{\dfrac{2}{5}-5}{\dfrac{2}{5}}\)

\(=\dfrac{-\dfrac{23}{5}}{\dfrac{2}{5}}\)

\(=-\dfrac{23}{2}\)

2) ĐKXĐ: \(x\ne9;x\ge0\)

\(A=\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{x+9\sqrt{x}}{9-x}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{x-9}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

3) \(P=A.B\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}.\dfrac{\sqrt{x}-5}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)

\(=\dfrac{\sqrt{x}+3-8}{\sqrt{x}+3}\)

\(=1-\dfrac{2}{\sqrt{x}+3}\)

Để P nhỏ nhất thì \(\dfrac{8}{\sqrt{x}+3}\) lớn nhất

Ta có:

\(\dfrac{8}{\sqrt{x}+3}\ge\dfrac{8}{3}\)

\(\Rightarrow P\) nhỏ nhất là \(1-\dfrac{8}{3}=-\dfrac{5}{3}\) khi \(x=0\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3}{2}\\x_1x_2=-\dfrac{7}{2}\end{matrix}\right.\)

\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\dfrac{37}{4}\)

\(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=\dfrac{153}{8}\)

\(C=x_1^4+x_2^4=\left(x_1^2+x_2^2\right)^2-2\left(x_1x_2\right)^2=\dfrac{977}{16}\)

\(D=\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\dfrac{\sqrt{65}}{2}\)

\(E=\left(2x_1+x_2\right)\left(2x_2+x_1\right)=2\left(x_1^2+x_2^2\right)+5x_1x_2=1\)

`3x^2+10x+3=0`

Ptr có: `\Delta'=5^2-3.3=16 > 0`

   `=>` Ptr có `2` nghiệm pb

 `=>` Áp dụng Viét có: `{(x_1+x_2=[-b]/a=-10/3),(x_1 .x_2=c/a=1):}`

~~~~~~~~~~~~~

`A=x_1 ^2+x_2 ^2`

`A=(x_1+x_2)^2-2x_1 .x_2`

`A=(-10/3)^2-2.1=82/9`

_______________________________________________________

`B=x_1 ^3+x_2 ^3`

`B=(x_1+x_2)(x_1 ^2-x_1 .x_2+x_2 ^2)`

`B=(x_1+x_2)[(x_1+x_2)^2 -3x_1 .x_2]`

`B=(-10/3).[(-10/3)^2-3.1]=-730/27`

_______________________________________________________

`C=x_1 ^4+x_2 ^4`

`C=(x_1 ^2+x_2 ^2)^2 -2x_1 ^2 .x_2 ^2`

`C=[(x_1+x_2)^2-2x_1 .x_2]^2-2(x_1 .x_2)^2`

`C=[(-10/3)^2-2.1]^2-2. 1^2=6562/81`

_______________________________________________________

`D=|x_1-x_2|`

`D=\sqrt{(x_1-x_2)^2}`

`D=\sqrt{(x_1+x_2)^2-4x_1.x_2}`

`D=\sqrt{(-10/3)^2-4.1}=8/3`

_______________________________________________________

`E=(2x_1+x_2)(2x_2+x_1)`

`E=4x_1 .x_2+2x_1 ^2+2x_2 ^2+x_1 .x_2`

`E=5x_1 . x_2+2(x_1+x_2)^2-4x_1 .x_2`

`E=x_1 .x_2+2(x_1+x_2)^2`

`E=1+2(-10/3)^2=209/9`

`a)` Thay `x=-3` vào ptr có:

   `(-3)^2-6.(-3)+2m+1=0`

`<=>9+18+2m+1=0`

`<=>m=-14`

`b)` Ptr có: `\Delta'=(-3)^2-(2m+1)=9-2m-1=8-2m`

Ptr có `2` nghiệm phân biệt `<=>\Delta' > 0`

    `<=>8-2m > 0<=>m < 4`

`c)` Ptr có nghiệm kép `<=>\Delta' =0`

           `<=>8-2m=0<=>m=4`

`a,` Đthang đi qua `A(3, 12)`.

`-> x = 3, y = 12 in y`.

`<=> 12 = 9a.`

`<=> a = 12/9 = 4/3.`

`b,` Đthang đi qua `B(-2;3)`.

`=> x = -2, y = 3 in y`.

`<=> 3=4a`.

`<=> a = 3/4`.

\(A=\dfrac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}>=1>0\)

=>A>|A|

Ta có: A= \(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)= \(1+\dfrac{1}{\sqrt{x}+1}\)

Vì x ≥0⇒\(\sqrt{x}\) ≥0⇒\(\sqrt{x}+1 \)≥ 1 ⇒ \(1+\dfrac{1}{\sqrt{x}+1}\)≥ 2

hay A≥ 2>0

Khi đó ta có: A=|A|

Vậy A=|A|