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

\(\Leftrightarrow x^2+6x+9-x^2+x=2\)

\(\Leftrightarrow7x+9=2\)

\(\Leftrightarrow7x=2-9\)

\(\Leftrightarrow7x=-7\)

\(\Leftrightarrow x=\dfrac{-7}{7}=-1\)

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

\(\Leftrightarrow4x^2+12x+9-\left(4x^2-3x+4x-3\right)=-1\)

\(\Leftrightarrow4x^2+12x+9-4x^2+3x-4x+3=-1\)

\(\Leftrightarrow11x+12=-1\)

\(\Leftrightarrow11x=-13\)

\(\Leftrightarrow x=\dfrac{-13}{11}\)

Đặt x² + 3x + 3 = a ;  x² - x - 1 = b ; -2x² - 2x - 1 = c ; -1 = d

Ta nhận thấy a³ + b³ + c³ + d³ = 0 (1) 

và a + b + c + d = 0

Khi đó ta có (1) <=>  (a + b)³ + (c + d)³ - 3ab(a + b) - 3cd(c + d) = 0

<=> ab(a + b) + cd(c + d) = 0

<=> (a + b)(ab - cd) = 0   

<=> \(\left[{}\begin{matrix}a=-b\\ab=cd\end{matrix}\right.\)

Với a = -b ta được x² + 3x + 3 = -x² + x + 1

<=> x² + x + 1 = 0 

<=> \(\left(x+\dfrac{1}{2}\right)^2=-\dfrac{3}{4}\)

=> Phương trình vô nghiệm

Với ab = cd 

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

\(\Leftrightarrow\) \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow\left(x^4+2x^3+x^2\right)-\left(4x^2+8x+4\right)=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-\left(2x+2\right)^2=0\)

\(\Leftrightarrow\left(x^2+3x+2\right).\left(x^2-x-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2.\left(x-2\right).\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)

x = -1

9x²+24x+16

Lời giải:

$(3x+4)^2=(3x)^2+2.3x.4+4^2=9x^2+24x+16$

P.s: Lần sau bạn chú ý ghi đầy đủ cả yêu cầu đề.

\(\text{∘}\) \(\text{Ans}\)

\(\downarrow\)

\(14x^2y^3-7xy^2\cdot\left(2x-3y\right)\)

`=`\(14x^2y^3-\left[7xy^2\cdot2x+7xy^2\cdot\left(-3y\right)\right]\)

`=`\(14x^2y^3-\left(14x^2y^2-21xy^3\right)\)

`=`\(14x^2y^3-14x^2y^2+21xy^3\)

\(\text{∘}\) \(\text{Kaizuu lv uuu.}\)

\(a,\left(4x-1\right)\left(x^2+12\right)\left(-x+4\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1>0\\x^2+12>0\left(LD\forall x\right)\\-x+4>0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x>1\\-x>-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{4}\\x< 4\end{matrix}\right.\)

Vậy \(S=\left\{x|\dfrac{1}{4}< x< 4\right\}\)

\(b,\left(2x-1\right)\left(5-2x\right)\left(1-x\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1< 0\\5-2x< 0\\1-x< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\)

Vậy \(S=\left\{x|1>x>\dfrac{5}{2}\right\}\)

Lời giải:

Áp dụng BĐT AM-GM:

$x^2+2^2\geq 4x$

$y^2+2^2\geq 4y$

$2(x^2+y^2)\geq 4xy$

$\Rightarrow 3(x^2+y^2)+8\geq 4(x+y+xy)=32$

$\Rightarrow x^2+y^2\geq 8$

Vậy $P_{\min}=8$ khi $x=y=2$

Lời giải:
$(\sqrt{2}x+\sqrt{8}y)^2=(\sqrt{2}x)^2+(\sqrt{8}y)^2+2\sqrt{2}x.\sqrt{8}y$

$=2x^2+8y^2+8xy$

Đề yêu cầu gì đó em?

Lời giải:
Gọi biểu thức trên là $A$

$4A=4a^2+4ab+4b^2-12a-12b+8064$

$=(4a^2+4ab+b^2)+3b^2-12a-12b+8064$

$=(2a+b)^2-6(2a+b)+(3b^2-6b)+8064$

$=(2a+b)^2-6(2a+b)+9+3(b^2-2b+1)+8052$

$=(2a+b-3)^2+3(b-1)^2+8052\geq 8052$

$\Rightarrow A\geq 2013$

Vậy $A_{\min}=2013$

\(a,\left(x-1\right)\left(x-2\right)>\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)-\left(x-1\right)^2>0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x-2\right)-\left(x-1\right)\right]>0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2-x+1\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(-1\right)>0\)

\(\Leftrightarrow x-1< 0\)

\(\Leftrightarrow x< 1\)

Vậy \(S=\left\{x|x< 1\right\}\)

\(b,\left(4x-1\right)\left(x^2+1\right)\left(-x+4\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1>0\\x^2+1>0\forall x\left(x^2\ge0\forall x\right)\\-x+4>0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{4}\\x< 4\end{matrix}\right.\)

Vậy \(S=\left\{x|\dfrac{1}{4}< x< 4\right\}\)