\(x^2+x+3=y^2\)

\(\Leftrightarrow4x^2+4x+12=4y^2\)

\(\Leftrightarrow\left(4x^2+4x+1\right)+11=4y^2\)

\(\Leftrightarrow\left(2x+1\right)^2-4y^2=-11\)

\(\Leftrightarrow\left(2x+1-2y\right).\left(2x+1+2y\right)=-11\)

do \(x,y\in Z\Rightarrow2x+1-2y;2x+1+2y\inƯ\left(-11\right)=\left\{\pm1;\pm11\right\}\)

ta có bẳng sau:

vậy các cặp nghiệm (x;y) của phương trình là:\(\left(x;y\right)\in\left\{\left(2;-3\right),\left(2;3\right),\left(-3;3\right),\left(-3;-3\right)\right\}\)

\(x^2+x+3=y^2\)

\(\Rightarrow4x^2+4x+12=4y^2\)

\(\Leftrightarrow\left(2x+1\right)^2+11=\left(2y\right)^2\)

\(\Leftrightarrow\left(2y\right)^2-\left(2x+1\right)^2=11\)

\(\Leftrightarrow\left(2y-2x-1\right)\left(2y+2x+1\right)=11\)

Ta lập bảng.

Với các số dương x;y ta có:

\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

Áp dụng:

\(\Rightarrow P=\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{ab\left(a+b\right)+abc}+\dfrac{1}{bc\left(b+c\right)+abc}+\dfrac{a}{ca\left(c+a\right)+abc}\)

\(\Rightarrow P\le\dfrac{abc}{ab\left(a+b+c\right)}+\dfrac{abc}{bc\left(a+b+c\right)}+\dfrac{abc}{ca\left(a+b+c\right)}\)

\(\Rightarrow P\le\dfrac{c}{a+b+c}+\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(P_{max}=1\) khi \(a=b=c=1\)

m(m -3)x - 2(2x - 2) = m

(m² - 3m . x ) + (-4x - 4) = m

-4xm² + 12xm - 4x² - 4m² + 12m - 4x = m

-4x . (m² + 12m - x - m² + 12m) = m

-4x . [(m² - m²) + (12 + 12) - x] = m

-4x . (24 - x) = m

-96x + 4x² = m

x. (-96 + 4x) = m

(x + 4x) - 96 = m

5x - 96 = m

\(\rightarrow\)5x = 96 (1)

x = 19,2

\(\rightarrow\)5 . 19,2 - 96 = 0

m = 0

(do mình ko giỏi về mấy cái thể loại toán như này nên có thể làm sai mong bạn thông cảm)

gfvfvfvfvfvfvfv555

`Answer:`

`m-3xyz+5x^2-7xy+9=6x^2+xyz+2xy+3-y^2`

`<=>m=(6x^2+xyz+2xy+3-y^2)+(3xyz-5x^2+7xy-9)`

`<=>(xyz+3xyz)+(6x^2-5x^2)+(2xy+7xy)-y^2+(3-9)`

`<=>m=4xyz+x^2+9xy-y^2-6`

gfvfvfvfvfvfvfv555

gfvfvfvfvfvfvfv555

\(\left[\left(6x+8\right)\left(6x+6\right)\right]\left(6x+7\right)^2=72\)

\(\left(36x^2+84x+48\right)\left(36x^2+84x+49\right)-72=0\)

Đặt \(36x^2+84x+48=a\) ta đc:

\(a\left(a+1\right)-72=0\)

\(a^2-a+\frac{1}{4}=\frac{287}{4}\)

\(\left(a-\frac{1}{2}\right)^2=\frac{287}{4}\)

Đến đây bạn tính a rồi thay vào tính x là xog.

`Answer:`

`(6x+8)(6x+6)(6x+7)^2=72`

`<=>(36x^2+36x+48x+48)(36x^2+84x+49)=72`

`<=>(36x^2+84x+48)(36x^2+84x+49)=72`

Đặt `n=36x^2+84x+48`

`<=>n(n+1)=72`

`<=>n^2+n-72=0`

`<=>n^2+9n-8n-72=0`

`<=>n(n+9)-8(n+9)=0`

`<=>(n+9)(n-8)=0`

`<=>(36x^2+84x+57)(36x^2+84x+40)=0`

`<=>36x^2+84x+40=0`

`<=>9x^2+21x+10=0`

`<=>9x^2+15x+6x+10=0`

`<=>(3x+2)(3x+5)=0`

\(\Leftrightarrow\orbr{\begin{cases}3x+2=0\\3x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{5}{3}\end{cases}}}\)

\(2x^2+4x=19-3y^2\)

\(\Leftrightarrow2x^2+4x+2=21-3y^2\)

\(\Leftrightarrow2.\left(x+1\right)^2=3.\left(7-y^2\right)\)(*)

ta có \(VT\)  \(2.\left(x+1\right)^2\)chẵn nên vế phải \(3.\left(7-y^2\right)\) chẵn

3 lẻ\(\Rightarrow7-y^2\) chẵn \(\Rightarrow y\) lẻ(1)

lại có \(VT\)\(2.\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow VP:3.\left(7-y^2\right)\ge0\forall y\)

\(\Rightarrow y^2\le7\)(2)

từ 1 và 2\(\Rightarrow y^2=1\)(do y nguyên)

\(\Rightarrow y=\pm1\)

ta có: thay y² vào PT(*) TA CÓ:

\(2.\left(x+1\right)^2=3.\left(7-1\right)\)

\(\Rightarrow2.\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(x+1\right)^2=9\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)

vậy cặp nghiệm (x;y) của phương trình là:\(\left(x;y\right)\in\left\{\left(2;\pm1\right),\left(-4;\pm1\right)\right\}\)

tham khảo

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\(P=a+b+\frac{1}{2a}+\frac{2}{b}\)

\(=\left(\frac{1}{2a}+\frac{a}{2}\right)+\left(\frac{2}{b}+\frac{b}{2}\right)-\frac{a+b}{2}\)

\(=\left(\frac{1}{2a}+\frac{a}{2}\right)+\left(\frac{2}{b}+\frac{b}{2}\right)-\frac{3}{2}\)

AD bất đẳng thức cố si cho 2 số ta đc:

\(P=\left(\frac{1}{2a}+\frac{a}{2}\right)+\left(\frac{2}{b}+\frac{b}{2}\right)-\frac{3}{2}\ge2.\sqrt{\frac{1}{2a}.\frac{a}{2}}+2.\sqrt{\frac{2}{b}.\frac{b}{2}}-\frac{3}{2}\)

\(P\ge2.\sqrt{\frac{1}{4}}+2.\sqrt{1}-\frac{3}{2}=2.\frac{1}{2}+2.1-\frac{3}{2}=\frac{3}{2}\)

VẬY minP=\(\frac{3}{2}\)