giup minh cau nay voi .Please help me :DDDDDD
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![](https://rs.olm.vn/images/avt/0.png?1311)
|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!!
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3}{2}\\x_1x_2=-\dfrac{7}{2}\end{matrix}\right.\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\dfrac{37}{4}\)
\(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=\dfrac{153}{8}\)
\(C=x_1^4+x_2^4=\left(x_1^2+x_2^2\right)^2-2\left(x_1x_2\right)^2=\dfrac{977}{16}\)
\(D=\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\dfrac{\sqrt{65}}{2}\)
\(E=\left(2x_1+x_2\right)\left(2x_2+x_1\right)=2\left(x_1^2+x_2^2\right)+5x_1x_2=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`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`
![](https://rs.olm.vn/images/avt/0.png?1311)
`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`.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}>=1>0\)
=>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|