\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+4x+y^3+3=0\\x^2y^3+y=2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x^2+2\left(x^2y^3+y\right)+y^3+3=0\left(1\right)\\x^2y^3+y=2x\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x^2+2x^2y^3+2y+y^3+3=0\Leftrightarrow\left(y+1\right)\left(2x^2y^2-2x^2y+y^2+2x-y+3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\Rightarrow x=-1\\2x^2y^2-2x^2y+y^2+2x^2-y+3=0\left(3\right)\end{matrix}\right.\)

\(\left(3\right)\Leftrightarrow2x^2\left(y^2-y+2\right)+y^2-y+3=0\)

\(\Rightarrow a=y^2-y+2=\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)

\(\Delta=0-4.2\left(y^2-y+2\right)\left(y^2-y+3\right)=-8\left(y^2-y+2\right)\left(y^2-y+3\right)\)

\(y^2-y+3=\left(y-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\)

\(\Rightarrow\Delta=-8\left(y^2-y+2\right)\left(y^2-y+3\right)< 0\)

\(\Rightarrow\left(3\right)\) không tồn tại nghiệm (x;y) nào

do đó hpt có  nghiệm x=y=-1

Lời giải:
ĐKXĐ: $x\neq -1$

PT $\Leftrightarrow (x-\frac{x}{x+1})^2+4=\frac{5x^2}{x+1}$

$\Leftrightarrow (\frac{x^2}{x+1})^2+4=\frac{5x^2}{x+1}$
Đặt $\frac{x^2}{x+1}=a$ thì pt trở thành:
$a^2+4=5a$

$\Leftrightarrow (a-1)(a-4)=0$

$\Leftrightarrow a=1$ hoặc $a=4$

Nếu $a=1\Leftrightarrow \frac{x^2}{x+1}=1$

$\Rightarrow x^2-x-1=0$

$\Leftrightarrow x=\frac{1\pm \sqrt{5}}{2}$

Nếu $a=4\Leftrightarrow \frac{x^2}{x+1}=4$

$\Rightarrow x^2-4x-4=0$

$\Leftrightarrow x=2\pm 2\sqrt{2}$

Lời giải:
ĐKXĐ: $a>0; a\neq 1$

\(A=\left[\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}(\sqrt{a}+1)}\right]:\frac{\sqrt{a}+1}{a\sqrt{a}}\)

\(=(\sqrt{a}-\frac{1}{\sqrt{a}}).\frac{a\sqrt{a}}{\sqrt{a}+1}=\frac{a-1}{\sqrt{a}}.\frac{a\sqrt{a}}{\sqrt{a}+1}=\frac{(\sqrt{a}-1)(\sqrt{a}+1)}{\sqrt{a}}.\frac{a\sqrt{a}}{\sqrt{a}+1}=a(\sqrt{a}-1)\)

ta có \(\sqrt{\left(a+c\right)\left(a+b\right)}\ge a+\sqrt{bc}\left(1\right)\)

thật vậy \(\left(1\right)\Leftrightarrow\left(a+c\right)\left(a+b\right)\ge a^2+2a\sqrt{bc}+bc\)

\(\Leftrightarrow ab+ac\ge2a\sqrt{bc}\Leftrightarrow b+c\ge2\sqrt{bc}\)(đúng theo BĐT cosi)

cminh tương tự \(\Rightarrow\sqrt{\left(b+c\right)\left(b+a\right)}\ge b+\sqrt{ac};\sqrt{\left(c+a\right)\left(c+b\right)}\ge c+\sqrt{ab}\)

\(\Rightarrow\dfrac{a}{\sqrt{\left(a+c\right)\left(a+b\right)}}\le\dfrac{a}{a+\sqrt{bc}}=\dfrac{1}{1+\dfrac{\sqrt{bc}}{a}}\)

\(tt\Rightarrow P\le\dfrac{1}{1+\dfrac{\sqrt{bc}}{a}}+\dfrac{1}{1+\dfrac{\sqrt{ac}}{b}}+\dfrac{1}{1+\dfrac{\sqrt{ab}}{c}}\)

\(đặt\left(\dfrac{\sqrt{bc}}{a};\dfrac{\sqrt{ac}}{b};\dfrac{\sqrt{ab}}{c}\right)=\left(x;y;z\right)\Rightarrow xyz=1\)

\(\Rightarrow P\le\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\)

ta đi chứng minh \(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\le\dfrac{3}{2}\)

\(\Leftrightarrow2\left(y+1\right)\left(z+1\right)+2\left(x+1\right)\left(z+1\right)+2\left(x+1\right)\left(y+1\right)\le3\left(x+1\right)\left(y+1\right)\left(z+1\right)\)

\(\Leftrightarrow2xy+2xz+2yz+4x+4y+4z+6\le3xyz+3+3xy+3xz+3yz+3x+3y+3z\)

ủa đến đây theo cách làm bth đúng rồi mà sao không ra nhỉ bạn xem lại hộ mình giống bài  n ày mình từng làm r

https://hoc24.vn/vip/289470733648/page-12

Toán 9

Ta có: \(\dfrac{2}{x^2+y^2}=\dfrac{x^2+y^2+z^2}{x^2+y^2}=1+\dfrac{z^2}{x^2+y^2}\le1+\dfrac{z^2}{2xy}\)(bđt cosi)

CMTT: \(\dfrac{2}{y^2+z^2}\le1+\dfrac{x^2}{2yz}\); \(\dfrac{2}{z^2+x^2}\le1+\dfrac{y^2}{2xz}\)

=> \(\dfrac{2}{x^2+y^2}+\dfrac{2}{y^2+z^2}+\dfrac{2}{z^2+x^2}\le3+\dfrac{z^2}{2xy}+\dfrac{x^2}{2yz}+\dfrac{y^2}{2xz}=3+\dfrac{x^3+y^3+z^3}{2xyz}\) (Đpcm)

Ta có: 2x2+y2=x2+y2+z2x2+y2=1+z2x2+y2≤1+z22xy2x2+y2=x2+y2+z2x2+y2=1+z2x2+y2≤1+z22xy(bđt cosi)

CMTT: 2y2+z2≤1+x22yz2y2+z2≤1+x22yz; 2z2+x2≤1+y22xz2z2+x2≤1+y22xz

=> 2x2+y2+2y2+z2+2z2+x2≤3+z22xy+x22yz+y22xz=3+x3+y3+z32xyz2x2+y2+2y2+z2+2z2+x2≤3+z22xy+x22yz+y22xz=3+x3+y3+z32xyz ( Đpcm )

\(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2a^2+2b^2+6ab\)

\(tt\Rightarrow B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\right)=6\left(a+b+c+d\right)^2\le6\)

\(dấu"='\Leftrightarrow a=b=c=d=\dfrac{1}{4}\)

a) \(\sqrt{\left|1-\sqrt{3}\right|^2}\cdot\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\sqrt{3+2\sqrt{3}+1}\)

\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)

\(=\left(\sqrt{3}\right)^2-1^2=3-1=2\)

b) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\sqrt{40\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\cdot4\sqrt{3}}\)

\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)

a) √∣∣1−√3∣∣2⋅√4+2√3|1−3|2⋅4+23

=√(√3−1)2⋅√3+2√3+1=(3−1)2⋅3+23+1

=(√3−1)√(√3+1)2=(3−1)(3+1)2

=(√3−1)(√3+1)=(3−1)(3+1)

=(√3)2−12=3−1=2=(3)2−12=3−1=2

b) 2√40√12−2√√75−3√5√4824012−275−3548

=2√40⋅2√3−2√5√3−3√5⋅4√3=240⋅23−253−35⋅43

=2√80√3−2√5√3−3√20√3=2803−253−3203

=8√5√3−2√5√3−6√5√3=0

Lời giải:

Bổ sung đk $m$ nguyên
Để pt có 2 nghiệm nguyên thì:

\(\Delta=m^2-4(m+2)=t^2\) với $t\in\mathbb{N}^*$

$\Leftrightarrow m^2-4m-8=t^2$

$\Leftrightarrow (m-2)^2-12=t^2$
$\Leftrightarrow 12=(m-2)^2-t^2=(m-2-t)(m-2+t)$

Vì $m-2-t, m-2+t$ có cùng tính chẵn lẻ nên $(m-2-t, m-2+t)=(2,6), (6,2), (-2,-6), (-6,-2)$

$\Rightarrow m=-2$ hoặc $m=6$

Thử lại thấy tm

Lời giải:

a. $\sqrt{x^2}-x+1=|x|-x+1=x-x+1=1$

b. $\sqrt{x^2}+2x=|x|+2x=-x+2x=x$

c. $\sqrt{(\frac{x}{y})^2}=|\frac{x}{y}|=\frac{x}{y}$ do $\frac{x}{y}\geq 0$ với $x\geq 0$ và $y>0$

d.

$\sqrt{(\frac{x}{y})^2}=|\frac{x}{y}|=\frac{-x}{y}$ do $\frac{x}{y}<0$ với $x>0; y<0$