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Ta có : (7x - 5y)²⁰¹⁸ + (3x - 2z)²⁰²⁰ + (xy + yz + xz - 4500)²⁰¹⁸ = 0

Ta có : \(\hept{\begin{cases}\left(7x-5y\right)^{2018}\ge0\\\left(3x-2z\right)^{2020}\ge0\\\left(xy+yz+xz-4500\right)^{2018}\ge0\end{cases}}\)

 \(\Rightarrow\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+xz-4500\right)^{2018}\ge0\)

Dấu bằng xảy ra <=> 

\(\begin{cases}7x=5y\\3x=2z\\xy+yz+xz=4500\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{y}{7}\\\frac{x}{2}=\frac{z}{3}\\xy+yz+xz=4500\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{10}=\frac{y}{14}\\\frac{x}{10}=\frac{z}{15}\\xy+yz+xz=4500\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=\frac{y}{14}=\frac{z}{15}\\x+y+z=4500\end{cases}}\)

Đặt \(\frac{x}{10}=\frac{y}{14}=\frac{z}{15}=k\Rightarrow\hept{\begin{cases}x=10k\\y=14k\\z=15k\end{cases}}\)

=> xy + yz + xz = 4500

<=> 10k.14k + 14k.15k + 10k.15k = 4500

=> 140.k² + 210.k² + 150.k² = 4500

=> k².(140 + 210 + 150) = 4500

=> k² . 500 = 4500

=> k² = 9

=> k = \(\pm3\)

Nếu k = 3

=> \(\hept{\begin{cases}x=30\\y=42\\z=45\end{cases}}\)

Nếu k = - 3

=> \(\hept{\begin{cases}x=-30\\y=-42\\z=-45\end{cases}}\)

bạn viêta đề rõ hơn đi

a ) x.y+14+2y+7x=-5

b) x.y+x+y=2

c) x.y-1=3x+5y+4

2 tìm x thuộc Z để A đạt giá trị nhỏ nhất

a) A=lxl+5

b) A=lx-5l-2018

l l là giá trị tuyệt đối nh

Lời giải:

Ta thấy:

$(7x-5y)^{2018}\geq 0, \forall x,y$

$(3x-2z)^{2020}\geq 0, \forall x,z$

$(xy+yz+xz-4500)^{2022}\geq 0, \forall x,y,z$

Do đó để tổng $(7x-5y)^{2018}+(3x-2z)^{2020}+(xy+yz+xz-4500)^{2022}=0$ thì:

$(7x-5y)^{2018}=(3x-2z)^{2020}=(xy+yz+xz-4500)^{2022}=0$

$\Leftrightarrow$ \(\left\{\begin{matrix} 7x=5y(1)\\ 3x=2z(2)\\ xy+yz+xz=4500(3)\end{matrix}\right.\)

Từ $(1);(2)\Rightarrow y=\frac{7}{5}x; z=\frac{3}{2}x$

Thay vào $(3)$:

$x.\frac{7}{5}x+\frac{7}{5}x.\frac{3}{2}x+x.\frac{3}{2}x=4500$

$\Leftrightarrow x^2=900\Rightarrow x=\pm 30$

Nếu $x=30\Rightarrow y=42; z=45$

Nếu $x=-30\Rightarrow y=-42; z=-45$

!

Cách khác:

\(\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+zx-4500\right)^{2022}=0\)

Ta có:

\(\left\{{}\begin{matrix}\left(7x-5y\right)^{2018}\ge0\\\left(3x-2z\right)^{2020}\ge0\\\left(xy+yz+zx-4500\right)^{2022}\ge0\end{matrix}\right.\forall x,y,z.\)

\(\Rightarrow\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+zx-4500\right)^{2022}\ge0\) \(\forall x,y,z.\)

\(\Rightarrow\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+zx-4500\right)^{2022}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(7x-5y\right)^{2018}=0\\\left(3x-2z\right)^{2020}=0\\\left(xy+yz+zx-4500\right)^{2022}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}7x-5y=0\\3x-2z=0\\xy+yz+zx-4500=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}7x=5y\\3x=2z\\xy+yz+zx=4500\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{x}{5}=\frac{y}{7}\\\frac{x}{2}=\frac{z}{3}\\xy+yz+zx=4500\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{10}=\frac{y}{14}\\\frac{x}{10}=\frac{z}{15}\\xy+yz+zx=4500\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{x}{10}=\frac{y}{14}=\frac{z}{15}\\xy+yz+zx=4500\end{matrix}\right.\)

Đặt \(\frac{x}{10}=\frac{y}{14}=\frac{z}{15}=k\Rightarrow\left\{{}\begin{matrix}x=10k\\y=14k\\z=15k\end{matrix}\right.\)

Có: \(xy+yz+zx=4500\)

\(\Rightarrow10k.14k+14k.15k+15k.10k=4500\)

\(\Rightarrow140.k^2+210.k^2+150.k^2=4500\)

\(\Rightarrow k^2.\left(140+210+150\right)=4500\)

\(\Rightarrow k^2.500=4500\)

\(\Rightarrow k^2=4500:500\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k=\pm3.\)

+ TH1: \(k=3.\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.3=30\\y=14.3=42\\z=15.3=45\end{matrix}\right.\)

+ TH2: \(k=-3.\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.\left(-3\right)=-30\\y=14.\left(-3\right)=-42\\z=15.\left(-3\right)=-45\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(30;42;45\right),\left(-30;-42;-45\right).\)

Chúc bạn học tốt!

a)7x³+3x⁴-x+5x²-6x³-2x⁴+2018+x³ = x⁴+2x³+5x²-x+2018

b)Hệ số cao nhất là 2018

Hệ số tự do là x

a) \(\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)

<=> \(9x^2-9x+2=9x^2+6x+1\)

<=>  \(15x=1\) <=> \(x=\frac{1}{15}\)

b) \(\left(4x-1\right)\left(x+1\right)=\left(2x-3\right)^2\)

<=> \(4x^2+3x-1=4x^2-12x+9\)

<=> \(15x^2=10\) <=> \(x=\frac{2}{3}\)

c) \(\left(5x+1\right)^2=\left(7x-3\right)\left(7x+2\right)\) <=> \(25x^2+10x+1=49x^2-7x-6\)

<=> \(24x^2-17x-7=0\) <=> \(24x^2-24x+7x-7=0\)

<=> \(\left(24x+7\right)\left(x-1\right)=0\) <=> \(\orbr{\begin{cases}x=-\frac{7}{24}\\x=1\end{cases}}\)

d) (4 - 3x)(4 + 3x) = (9x - 3)(1 - x)

<=> 16 - 9x² = 12x - 9x2 - 3

<=> 12x = 19

<=> x = 19/12

e) x(x + 1)(x + 2)(x + 3) = 24

<=> (x² + 3x)(x² + 3x + 2) = 24

<=> (x² + 3x)²  + 2(x² + 3x) - 24 = 0

<=> (x² + 3x)² + 6(x² + 3x) - 4(x² + 3x) - 24 = 0

<=> (x² + 3x + 6)(x² + 3x - 4) = 0

<=> \(\orbr{\begin{cases}x^2+3x+6=0\\x^2+3x-4=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

g) (7x - 2)² = (7x - 3)(7x + 2)

<=> 49x² - 28x + 4 = 49x² - 7x - 6

<=> 21x = 10 <=> x = 10/21