\(\left\{{}\begin{matrix}1+a+b+c=0\\8+4a+2b+c=0\\a+b=-16\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=-1\\4a+2b+c=-8\\a+b=-16\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{9}{2}\\b=-\dfrac{41}{2}\\c=15\end{matrix}\right.\)

\(\dfrac{x+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\) (1)

Ta có: \(4x^3-3=29\)

\(\Rightarrow4x^3=32\Rightarrow x^3=8\)

\(\Rightarrow x=2\)

Thay \(x=2\) vào điều (1) ta có:

\(\dfrac{2+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\)

\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=\dfrac{18}{9}\)

\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=2\)

\(\Rightarrow\left\{{}\begin{matrix}y-25=2.\left(-16\right)\\z+49=2.25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y-25=-32\\z+49=50\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=-7\\z=1\end{matrix}\right.\)

Vậy giá trị của biểu thức \(A=x+2y+3z\) là:

\(A=2+2.\left(-7\right)+3.1=2-14+3=-9\)

Chúc bạn học tốt!!!

Ta có : \(4x^3-3=29\)

\(\Rightarrow4x^3=32\)

\(\Rightarrow x^3=8\)

\(\Rightarrow x=2\)

Thay x = 2 vào \(\dfrac{x+16}{9}=\dfrac{y-25}{-16}\) ta có :

\(\dfrac{2+16}{9}=\dfrac{y-25}{-16}\)

\(\Rightarrow2=\dfrac{y-25}{-16}\)

\(\Rightarrow y-25=-32\)

\(\Rightarrow y=-7\)

Thay \(y=-7\) vào \(\dfrac{y-25}{-16}=\dfrac{z+49}{25}\) ta có :

\(\dfrac{-7-25}{-16}=\dfrac{z+49}{25}\)

\(\Rightarrow2=\dfrac{z+49}{25}\)

\(\Rightarrow z+49=50\)

\(\Rightarrow z=1\)

Thay x = 2; y = -7; z = 1 vào biểu thức A ta có :

\(A=2+2.\left(-7\right)+3.1\)

\(A=-9\)

Vậy A = -9

Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng

Nguyễn Huy Tú Lightning Farron Akai Haruma

bài 1 dễ mà bn .bn chỉ cần tính x rùi thay vào thui mà

2a Tacó

y(x-y).(-5)=x(x-y)

-5y^2+5xy=xx-xy

5xy+xy=x^2+5y^2

6xy=x^2-5y^2

2b xét 2 th

th1 cùng(-)

th2 cùng (+)

Bài 1: Vì: 2x^3 - 1 = 15
=> 2x^3 = 16
=> x^3 = 8
=> x = 2 (1)
Ta có:
* (x + 16)/9 = (y - 25)/16
<=> (2 + 16)/9 = (y - 25)/16
<=> 18/9 = (y - 25)/16
<=> 2 = (y - 25)/16
<=> y - 25 = 16.2 = 32
=> y = 32+25 = 57 (2)

* (x + 16)/9 = (z + 9)/25
<=> (2 + 16)/9 = (z + 9)/25
<=> 2 = (z + 9)/25
<=> z + 9 = 25.2 = 50
=> z = 50 - 9 = 41 (3)
Từ (1), (2) và (3) => x + y + z = 2 + 57 + 41 = 100

Bài 2:

c) vì a,b,c là độ dài các cạnh của tam giác:

\(\Rightarrow\left\{{}\begin{matrix}a< b+c\\b< a+c\\c< a+b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c}< 1\\\dfrac{b}{a+c}< 1\\\dfrac{c}{a+b}< 1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c}< \dfrac{2a}{a+b+c}\\\dfrac{b}{a+c}< \dfrac{2b}{a+b+c}\\\dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\end{matrix}\right.\)

\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}< \dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\) (đpcm)

Bài 3:

b)

Để \(\left(x-3\right)\left(x+\dfrac{1}{2}\right)>0\)

thì x-3 và \(x+\dfrac{1}{2}\) phải cùng dấu.

mà: \(x-3< x+\dfrac{1}{2}\forall x\)

\(\Rightarrow\left[{}\begin{matrix}0< x-3< x+\dfrac{1}{2}\\0>x+\dfrac{1}{2}>x-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>3\\x< -\dfrac{1}{2}\end{matrix}\right.\)

Vậy với \(\left[{}\begin{matrix}x>3\\x< -\dfrac{1}{2}\end{matrix}\right.\) thì \(\left(x-3\right)\left(x+\dfrac{1}{2}\right)>0\).

P(x)=\(ax^2+bx+c\) (1)(a\(\ne0\) )

Ta có : 

\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}\)\(\Rightarrow\left\{{}\begin{matrix}b=2a\\c=3a\end{matrix}\right.\)(2)

Thay(2) vào (1)\(\Rightarrow P\left(x\right)=ax^2+2ax+3a\)

\(\Rightarrow\dfrac{P\left(-2\right)-3P\left(-1\right)}{a}=\dfrac{4a-4a+3a-3\left(a-2a+3a\right)}{a}\)=\(\dfrac{3a-3a+6a-9a}{a}=\dfrac{-3a}{a}=-3\)

\(4x^3-3=29\Rightarrow x^3=\dfrac{29+3}{4}=8\Rightarrow x=\sqrt[3]{8}=2\)

Thay số: \(\dfrac{x+16}{9}=\dfrac{2+16}{9}=2\)

Suy ra: \(y=\left(-16\right)\cdot2+25\Leftrightarrow y=-7\) và \(z=25\cdot2-49\Leftrightarrow z=1\)

\(A=x+2y+3z\Leftrightarrow2+\left(-14\right)+3=-9\)

\(4x^3-3=29\Rightarrow x^3=\dfrac{32}{4}=2^3\Rightarrow x=3\)

\(\dfrac{19}{9}=\dfrac{2y-2.25}{-32}=\dfrac{3z+49.3}{75}=\dfrac{2y+3z+49.3-25.2}{75-32}=\dfrac{2y+3z+97}{43}\)

\(\dfrac{\left(2y+3z+3\right)+94}{43}=\dfrac{19}{9}\) \(\Rightarrow\left(x+2y+3z\right)=\dfrac{43.19}{9}-94\)

-9