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9 tháng 7 2019

Không chắc đâu:v

a) Ta luôn có \(\left(x-1\right)^2+\left(2x-y-3\right)^2+\left(y+z\right)^2\ge0\forall x,y,z\)

Để đẳng thức xảy ra tức là \(\left(x-1\right)^2+\left(2x-y-3\right)^2+\left(y+z\right)^2=0\) (theo đề bài)

Thì \(\left\{{}\begin{matrix}x=1\\y=2x-3=2.1-3=-1\\z=-y=1\end{matrix}\right.\)

Vậy...

b) Ta luôn có \(VT\ge0\) với mọi x, y. Mà theo đề bài \(VT\le0\)

Do vậy \(VT=0\Leftrightarrow\left(2x+3\right)^{1998}+\left(3y-5\right)^{2000}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\y=\frac{5}{3}\end{matrix}\right.\)

Bài này của lớp 10 ?? Hơi lạ....

Bài 1:

a) \(5x-15y=5\left(x-3y\right)\)

b) \(\dfrac{3}{5}x^2+5x^4-x^2y=x^2\left(\dfrac{3}{5}+5x^2-y\right)\)

c) \(14x^2y^2-21xy^2+28x^2y=7xy\left(2xy-3y+4x\right)\)

d) \(\dfrac{2}{7}x\left(3y-1\right)-\dfrac{2}{7}y\left(3y-1\right)=\dfrac{2}{7}\left(3y-1\right)\left(x-y\right)\)

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

f) \(\left(x+y\right)^2-4x^2=\left(-x+y\right)\left(3x+y\right)\)

g) \(27x^3+\dfrac{1}{8}=\left(3x+\dfrac{1}{2}\right)\left(6x^2+1,5x+\dfrac{1}{4}\right)\)

h) \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=6x^2y+2y^3=2y\left(3x^2+y\right)\)

Bài 2:

a) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\Rightarrow x=-1\\x+2=0\Rightarrow x=-2\end{matrix}\right.\)

b) \(x\left(3x-2\right)-5\left(2-3x\right)=0\)

\(\Rightarrow x\left(3x-2\right)+5\left(3x-2\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=0\Rightarrow x=\dfrac{2}{3}\\x+5=0\Rightarrow x=-5\end{matrix}\right.\)

c) \(\dfrac{4}{9}-25x^2=0\)

\(\Rightarrow\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\Rightarrow x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0\Rightarrow x=\dfrac{-2}{15}\end{matrix}\right.\)

d) Có tới 2 dấu "=".

28 tháng 9 2017

bài 1 dễ mk ko lm nữa nhé

bafi2:

a,x(x+1)(x+2)=0

x=0 ; x=-1 ; x=-2

b,x(3x-2)+5(3x-2)=0

(x+5)(3x-2)=0

x=-5 ; x=2/3

c,

(2/3)2- (5x)2=0

(2/3-5x)(2/3+5x)=0

x=+-2/15

d, X2-2*1/2x+(1/2)2=0

(X-1/2)22=0

X=1/2

4 tháng 8 2019

Bài 1

d, \(x^2+2xy+y^2-2x-2y+1\)

\(\Rightarrow x^2+y^2=1+2xy-2y-2x\)

\(\Rightarrow\left(x+y-1\right)^2\)

Bài 2:

a, \(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)

\(\Leftrightarrow x^2+2x+1=x^2=5x+2x+10\)

\(\Leftrightarrow-5x=9\)

\(\Leftrightarrow x=-\frac{9}{5}\)

b,\(\left(x+3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

c, \(4x^2-9=0\)

\(\Leftrightarrow4x^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\\frac{3}{2}\end{matrix}\right.\)

d,\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)

\(\Leftrightarrow16x^2-40x+25-\left(9x^2-24x+16\right)=0\)

\(\Leftrightarrow16x^2-40x+25-9x^2+24x-16=0\)

\(\Leftrightarrow7x^2-16x+9=0\)

\(\Leftrightarrow x=\frac{-\left(-16\right)\pm\sqrt{\left(-16\right)^2-4.7.9}}{14}\)

\(\Leftrightarrow x=\frac{16\pm\sqrt{256-252}}{14}\)

\(\Leftrightarrow x=\frac{16\pm\sqrt{4}}{14}\)

\(\Leftrightarrow x=\frac{16\pm2}{14}\)

\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{16+2}{14}\\\frac{16-2}{14}\end{matrix}\right.\)

\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{9}{7}\\1\end{matrix}\right.\)

4 tháng 8 2019

1.a)\(3x-3y+x^2-2xy+y^2\)

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

\(=\left(x-y\right)\left(3+x-y\right)\)

d)\(x^2+2xy+y^2-2x-2y+1\)

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

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

2.a)\(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)

\(\Leftrightarrow x^2+2x+1-x^2-7x-10=0\)

\(\Leftrightarrow-5x-9=0\)

\(\Leftrightarrow-5x=9\)

\(\Leftrightarrow x=-\frac{9}{5}\). Vậy \(S=\left\{-\frac{9}{5}\right\}\)

b)\(\left(x+3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\).Vậy \(S=\left\{-3;-5\right\}\)

c)\(4x^2-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{3}{2}\end{matrix}\right.\). Vậy \(S=\left\{\pm\frac{3}{2}\right\}\)

d)\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)

\(\Leftrightarrow\left(4x-5+3x-4\right)\left(4x-5-3x+4\right)=0\)

\(\Leftrightarrow\left(7x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{7}\\x=1\end{matrix}\right.\). Vậy \(S=\left\{1;\frac{9}{7}\right\}\)

3.Ta có:

8x^2-26x+m 2x-3 4x-7 -14x+m m+21

Để \(A\left(x\right)⋮B\left(x\right)\) thì: \(m+21⋮2x-3\)

\(\Rightarrow m+21=0\)

\(\Rightarrow m=-21\)

Vậy...!

NV
13 tháng 3 2020

a/ - Với \(x>\frac{1}{4}\) PT vô nghiêm

- Với \(x\le\frac{1}{4}\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x^2+4x-2=0\\x^2-4x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2+\sqrt{6}\left(l\right)\\x=-2-\sqrt{6}\\x=4\left(l\right)\\x=0\end{matrix}\right.\)

2.

- Với \(x\ge-\frac{1}{4}\Leftrightarrow4x+1=x^2+2x-4\)

\(\Leftrightarrow x^2-2x-5=0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{6}\\x=1-\sqrt{6}\left(l\right)\end{matrix}\right.\)

- Với \(x< -\frac{1}{4}\)

\(\Leftrightarrow-4x-1=x^2+2x-4\)

\(\Leftrightarrow x^2+6x-3=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3+2\sqrt{3}\left(l\right)\\x=-3-2\sqrt{3}\end{matrix}\right.\)

NV
13 tháng 3 2020

3.

- Với \(x\ge\frac{5}{3}\)

\(\Leftrightarrow3x-5=2x^2+x-3\)

\(\Leftrightarrow2x^2-2x+2=0\left(vn\right)\)

- Với \(x< \frac{5}{3}\)

\(\Leftrightarrow5-3x=2x^2+x-3\)

\(\Leftrightarrow2x^2+4x-8=0\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)

4. Do hai vế của pt đều không âm, bình phương 2 vế:

\(\Leftrightarrow\left(x^2-2x+8\right)^2=\left(x^2-1\right)^2\)

\(\Leftrightarrow\left(x^2-2x+8\right)^2-\left(x^2-1\right)^2=0\)

\(\Leftrightarrow\left(2x^2-2x+7\right)\left(-2x+9\right)=0\)

\(\Leftrightarrow-2x+9=0\Rightarrow x=\frac{9}{2}\)

NV
13 tháng 3 2020

a/ Đặt \(\left|x\right|=t\ge0\Rightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\end{matrix}\right.\)

\(\Rightarrow\left|x\right|=2\Rightarrow x=\pm2\)

b/ \(\Leftrightarrow\left(x+1\right)^2+\left|x+1\right|-6=0\)

Đặt \(\left|x+1\right|=t\ge0\Rightarrow t^2+t-6=0\Rightarrow\left[{}\begin{matrix}t=-3\left(l\right)\\t=2\end{matrix}\right.\)

\(\Rightarrow\left|x+1\right|=2\Rightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

c/ \(\Leftrightarrow\left(x+1\right)^2-5\left|x+1\right|+4=0\)

Đặt \(\left|x+1\right|=t\ge0\Rightarrow t^2-5t+4=0\Rightarrow\left[{}\begin{matrix}t=1\\t=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+1\right|=1\\\left|x+1\right|=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=4\\x+1=-4\end{matrix}\right.\)

NV
13 tháng 3 2020

d. \(\Leftrightarrow\left(x-1\right)^2+5\left|x-1\right|+4=0\)

Đặt \(\left|x+1\right|=t\ge0\Rightarrow t^2+5t+4=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=-4\left(l\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

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

Đặt \(\left|x-2\right|=t\ge0\)

\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

f. \(\Leftrightarrow\left(2x-5\right)^2+4\left|2x-5\right|-12=0\)

Đặt \(\left|2x-5\right|=t\ge0\)

\(\Rightarrow t^2+4t-12=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-6\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left|2x-5\right|=2\Rightarrow\left[{}\begin{matrix}2x-5=2\\2x-5=-2\end{matrix}\right.\)

16 tháng 8 2016

1)Thấy: x=0;y=0 không phải là nghiệm của hệ.

\(\begin{cases}x^3-8x=y^3+2y\\x^2-3=3\left(y^2+1\right)\end{cases}\)

\(\Leftrightarrow\begin{cases}x^3-8x=y^3+2y\\x^2=3\left(y^2+2\right)\end{cases}\)

\(\Leftrightarrow\begin{cases}x^3-8x=y\left(y^2+2\right)\\x^2y=3y\left(y^2+2\right)\end{cases}\)

Trừ vế theo vế hai phương trình,đc:

\(x^3-8x-\frac{x^2y}{3}=0\Leftrightarrow y=\frac{3\left(x^3-8x\right)}{x^2}\)

\(\Leftrightarrow y=\frac{3\left(x^2-8\right)}{x}\).Thay \(y=\frac{3\left(x^2-8\right)}{x}\) vào pt 2 đc:

\(26x^4-426x^2-1728=0\)

\(\Leftrightarrow\begin{cases}x^2=9\\x^2=\frac{96}{13}\end{cases}\) dễ nhé oaoa

 

16 tháng 8 2016

lần sau bn đăng ít 1 thôi nhé

AH
Akai Haruma
Giáo viên
28 tháng 2 2020

Bài 1:

Ta có:

\(x^2+xy+y^2=\frac{3}{4}(x^2+2xy+y^2)+\frac{1}{4}(x^2-2xy+y^2)\)

\(=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\geq \frac{3}{4}(x+y)^2\)

\(\Rightarrow \sqrt{x^2+xy+y^2}\geq \frac{\sqrt{3}(x+y)}{2}\)

Hoàn toàn tương tự:

\(\sqrt{y^2+yz+z^2}\geq \frac{\sqrt{3}(y+z)}{2}; \sqrt{z^2+xz+x^2}\geq \frac{\sqrt{3}(x+z)}{2}\)

Cộng theo vế các BĐT trên:

\(\Rightarrow \sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+xz+x^2}\geq \sqrt{3}(x+y+z)\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z$

AH
Akai Haruma
Giáo viên
28 tháng 2 2020

Bài 2:

BĐT cần chứng minh tương đương với:

$4(a^9+b^9)-(a+b)(a^3+b^3)(a^5+b^5)\geq 0$

$\Leftrightarrow 4(a+b)(a^8-a^7b+a^6b^2-a^5b^3+a^4b^4-a^3b^5+a^2b^6-ab^7+b^8)-(a+b)(a^8+a^3b^5+a^5b^3+b^8)\geq 0$

$\Leftrightarrow 4(a^8-a^7b+a^6b^2-a^5b^3+a^4b^4-a^3b^5+a^2b^6-ab^7+b^8)-(a^8+a^3b^5+a^5b^3+b^8)\geq 0$

$\Leftrightarrow 3a^8+3b^8+4a^6b^2+4a^2b^6+4a^4b^4-(4a^7b+4ab^7+5a^5b^3+5a^3b^5)\geq 0$

$\Leftrightarrow (a-b)^2(a^2-ab+b^2)(3a^4+5a^3b+7a^2b^2+5ab^3+3b^4)\geq 0$

BĐT trên luôn đúng vì:

$(a-b)^2\geq 0, \forall a,b$

$a^2-ab+b^2=(a-\frac{b}{2})^2+\frac{3}{4}b^2\geq 0, \forall a,b$

$3a^4+5a^3b+7a^2b^2+5ab^3+3b^4=3(a^4+b^4+2a^2b^2)+a^2b^2+5ab(a^2+b^2)$

$=3(a^2+b^2)^2+5ab(a^2+b^2)+a^2b^2$

$=(a^2+b^2)(3a^2+3b^2+5ab)+a^2b^2=(a^2+b^2)[3(a+\frac{5}{6}b)^2+\frac{11}{12}b^2]+a^2b^2\geq 0$ với mọi $a,b$

Do đó ta có đpcm.

Dấu "=" xảy ra khi $a=b$ hoặc $a+b=0$

5 tháng 2 2017

a) x3+4x2+x-6=0

<=> x3+3x2+x2+3x-2x-6=0

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

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

<=> \(\left[\begin{matrix}x+3=0\\x^2+x-2=0\end{matrix}\right.\)<=> \(\left[\begin{matrix}x=-3\\\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\end{matrix}\right.\)

<=> \(\left[\begin{matrix}x=-3\\x=1\\x=-2\end{matrix}\right.\)

Vậy ...

b) x3-3x2+4=0

<=> x3-2x2-x2+4=0

<=> x2(x-2)-(x-2)(x+2)=0

<=> (x-2)(x2-x-2)=0

<=> \(\left[\begin{matrix}x-2=0\\x^2-x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=2\\\left(x-\frac{1}{2}\right)^2=\frac{9}{4}\end{matrix}\right.\)

<=> \(\left[\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Vậy ...

5 tháng 2 2017

c) x4+2x3+2x2-2x-3=0

<=> x4+x3+x3+x2+x2+x-3x-3=0

<=> x3(x+1)+x2(x+1)+x(x+1)-3(x+1)=0

<=> (x+1)(x3+x2+x-3)=0

<=> (x+1)(x3-x2+2x2-2x+3x-3)=0

<=> (x+1)[x2(x-1)+2x(x-1)+3(x-1)]=0

<=> (x+1)(x-1)(x2+2x+3)=0

Mà x2+2x+3=x2+2x+1+2=(x+1)2+2>0

<=> (x+1)(x-1)=0

<=>\(\left[\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\)<=> \(\left[\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy ...