1. b3+b= 3                                       

(b3+b)=3                            

b.(3+1)=3

b. 4= 3

b=\(\dfrac{3}{4}\)

a3+a= 3                                       b3

(a3+a)=3                            

a.(3+1)=3

a. 4= 3

a=\(\dfrac{3}{4}\)

2

a: \(a^2+4a=b^2+4b+1\)

=>\(a^2+4a-b^2-4b=0\)

=>(a-b)(a+b)+4(a-b)=0

=>(a-b)(a+b+4)=0

mà a-b<>0

nên a+b+4=0

=>a+b=-4

b: Đặt \(X=a^3+b^3\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=\left(-4\right)^3-3ab\cdot\left(-4\right)=-64+12ab\)

\(a^2+4a=1\)

=>\(a^2+4a-1=0\)

=>\(a^2+4a+4-5=0\)

=>\(\left(a+2\right)^2=5\)

=>\(\left[\begin{array}{l}a+2=\sqrt5\\ a+2=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}a=\sqrt5-2\\ a=-\sqrt5-2\end{array}\right.\)

\(b^2+4b=1\)

=>\(b^2+4b-1=0\)

=>\(b^2+4b+4-5=0\)

=>\(\left(b+2\right)^2=5\)

=>\(\left[\begin{array}{l}b+2=\sqrt5\\ b+2=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}b=\sqrt5-2\\ b=-\sqrt5-2\end{array}\right.\)

Vì a<>b nên sẽ có hai trường hợp sau:

TH1: \(a=\sqrt5-2;b=-\sqrt5-2\)

=>\(ab=\left(\sqrt5-2\right)\left(-\sqrt5-2\right)=-\left(\sqrt5-2\right)\left(\sqrt5+2\right)=-1\)

X=-64+12ab

=-64-12

=-76

TH2: \(a=-\sqrt5-2;b=\sqrt5-2\)

=>\(ab=\left(\sqrt5-2\right)\left(-\sqrt5-2\right)=-\left(\sqrt5-2\right)\left(\sqrt5+2\right)=-1\)

X=-64+12ab

=-64-12

=-76

Vậy: X=-76

c: Đặt \(Y=a^4+b^4\)

\(=\left(a^2+b^2\right)^2-2a^2b^2\)

\(=\left\lbrack\left(a+b\right)^2-2ab\right\rbrack^2-2\cdot\left(ab\right)^2\)

\(=\left\lbrack\left(-4\right)^2-2\cdot\left(-1\right)\right\rbrack^2-2\cdot\left(-1\right)^2=\left\lbrack16+2\right\rbrack^2-2\)

\(=18^2-2\)

=324-2

=322

Ta có \(-\dfrac{4ab^2}{4b^2+1}\ge-\dfrac{4ab^2}{2\sqrt{4b^2}}=\dfrac{4ab^2}{4b}=ab\)

\(-\dfrac{4a^2b}{4a^2+1}\ge-\dfrac{4a^2b}{2\sqrt{4a^2}}=\dfrac{4a^2b}{4a}=ab\)

Mà \(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}=\dfrac{a\left(4b^2+1\right)}{4b^2+1}-\dfrac{4ab^2}{4b^2+1}+\dfrac{b\left(4a^2+1\right)}{4a^2+1}-\dfrac{4ab^2}{4a^2+1}\ge a-ab+b-ab=4ab-2ab=2ab\)

Mà \(a+b=4ab\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=4\ge\dfrac{2}{2\sqrt{ab}}\Rightarrow4\sqrt{ab}\ge2\Rightarrow ab\ge\dfrac{1}{4}\)

\(\Rightarrow2ab\ge\dfrac{1}{2}\Rightarrow\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)

Dấu "=" \(\Leftrightarrow a=b=\dfrac{1}{2}\)

Lời giải:

ĐK $\Rightarrow \frac{1}{a}+\frac{1}{b}=4$

Đặt $\frac{1}{x}=a; \frac{1}{y}=b$ thì bài toán trở thành:

Cho $a,b>0$ thỏa mãn $a+b=4$. CMR:

$P=\frac{x^2}{y(x^2+4)}+\frac{y^2}{x(y^2+4)}\geq \frac{1}{2}$

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Áp dụng BĐT AM-GM:

$\frac{x^2}{y(x^2+4)}+\frac{y(x^2+4)}{64}\geq \frac{x}{4}$

$\frac{y^2}{x(y^2+4)}+\frac{x(y^2+4)}{64}\geq \frac{y}{4}$

Cộng theo vế và rút gọn:

$P\geq \frac{3(x+y)-xy}{16}=\frac{12-xy}{16}$

Mà $xy\leq \frac{(x+y)^2}{4}=4$

$\Rightarrow P\geq \frac{12-4}{16}=\frac{1}{2}$

Ta có đpcm.

\(a^3-a^2b+ab^2-6b^3=0\)

\(\Leftrightarrow\left(a-2b\right)\left(a^2+ab+3b^2\right)=0\left(1\right)\)

Vì a>b>0 =>a²+ab+3b²>0 nên từ (1) ta có a=2b

Vậy biểu thức \(A=\frac{a^4-4b^4}{b^4-4a^4}=\frac{16b^4-4b^4}{b^4-64b^4}=\frac{12b^4}{-63b^4}=-\frac{4}{21}\)

Ta đặt \(a^2+4b+3=k^2\) 

\(\Leftrightarrow k^2-a^2\equiv3\left[4\right]\)

Mà \(k^2,a^2\equiv0,1\left[4\right]\) nên \(k^2⋮4,a^2\equiv1\left[4\right]\) \(\Rightarrow k⋮2,a\equiv1\left[2\right]\)

Đặt \(k=2l,a=2c+1>b\), ta có \(\left(2c+1\right)^2+4b+3=4l^2\)

\(\Leftrightarrow4c^2+4c+4b+4=4l^2\)

\(\Leftrightarrow c^2+c+1+b=l^2\)

Nếu \(b< c\) thì \(c^2< c^2+c+1+b< c^2+2c+1=\left(c+1\right)^2\), vô lí.

Nếu \(c< b< 2c+1\) thì

\(\left(c+1\right)^2< c^2+c+1+b< c^2+4c+4=\left(c+2\right)^2\), cũng vô lí.

Do vậy, \(c=b\) hay \(a=2b+1\)

Từ đó \(b^2+4a+12=b^2+4\left(2b+1\right)+12\) \(=b^2+8b+16\) \(=\left(b+4\right)^2\) là SCP. Suy ra đpcm.

😁😁😁😁

Đáp án B