help me, please

1. a . 3x² - 6x = 0

\(\Leftrightarrow3x\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}3x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

b. x³ - 13x = 0

\(\Leftrightarrow x\left(x^2-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{13}\end{cases}}\)

c. 5x ( x - 2001 ) - x + 2001 = 0

<=> 5x ( x - 2001 ) - ( x - 2001 ) = 0

\(\Leftrightarrow\left(5x-1\right)\left(x-2001\right)=0\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-2001=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2001\end{cases}}\)

2. a. \(2x^2+4x-8=2\left(x+1\right)^2-10\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x+1\right)^2-10\ge-10\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy GTNN của bt trên = - 10 <=> x = - 1

b. \(-x^2-8x+1=-\left(x+4\right)^2+17\)

Vì \(\left(x+4\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+4\right)^2+17\le17\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+4\right)^2=0\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

Vậy GTLN của bt trên = 17 <=> x = - 4

tương tự baì đẳng trên mình vừa làm đấy

|A| <= 0 với mọi A

thì -|A| <= 0 vứi mọi A

tương tự với bình phương một số

a) \(A=5-8x-x^2=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left[\left(x+4\right)^2-21\right]\)

\(=-\left(x+4\right)^2+21\le21\)

Vậy \(A_{max}=21\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

\(B=5x-3x^2=-3\left(x^2-\frac{5}{3}x\right)\)

\(=-3\left(x^2-\frac{5}{3}x+\frac{35}{36}-\frac{25}{36}\right)\)

\(=-3\left[\left(x-\frac{5}{6}\right)^2-\frac{25}{36}\right]\)

\(=-3\left[\left(x-\frac{5}{6}\right)^2\right]+\frac{25}{12}\le\frac{25}{12}\)

Vậy \(B_{min}=\frac{25}{12}\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)

Bài 6:

a) \(x\left(x-2\right)+x-2=0\)

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

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

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

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

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

c) \(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)

\(\Leftrightarrow3x^2-15x-2x-3x^2+2+3x=30\)

\(\Leftrightarrow-14x+2=30\)

\(\Leftrightarrow-14x=28\)

\(\Leftrightarrow x=-2\)

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

\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)

\(\Leftrightarrow2x+16=0\)

\(\Leftrightarrow2x=-16\)

\(\Leftrightarrow x=-8\)

Em cần gấp bây h ạ :<

Bài 1:

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

b) \(x^4+x^3+2x^2+x+1\)

\(=x^4+x^2+x^3+x+x^2+1\)

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

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

Bài 1

a) \(A=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x^2+\frac{x}{2}-\frac{1}{2}\right)=2\left(x^2+2.\frac{1}{4}.x+\frac{1}{16}-\frac{9}{16}\right)\)\(=2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\)

Vì \(\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

Dấu "=" xảy ra khi \(\left(x+\frac{1}{4}\right)^2=0\Leftrightarrow x+\frac{1}{4}=0\Leftrightarrow x=-\frac{1}{4}\)

Vậy minA=-9/8 khi x=-1/4

b)\(B=4x^2-4xy+2y^2+1=\left(4x^2-4xy+y^2\right)+y^2+1=\left(2x-y\right)^2+y^2+1\)

Vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)=>\(\left(2x-y\right)^2+y^2\ge0\Rightarrow B=\left(2x-y\right)^2+y^2+1\ge1\)

Dấu "=" xảy ra khi (2x-y)²=y²=0 <=> 2x-y=y=0 <=> x=y=0

Vậy minB=1 khi x=y=0

lý luận tương tự bài 1, bài này mình làm tắt

Bài 2:

a) \(C=5x-3x^2+2=-\left(3x^2-5x-2\right)=-3\left(x^2-\frac{5}{3}x-\frac{2}{3}\right)\)

\(=-3\left(x^2-2.\frac{5}{6}.x+\frac{25}{35}-\frac{49}{36}\right)=-3\left[\left(x-\frac{5}{6}\right)^2-\frac{49}{36}\right]=\frac{49}{12}-3\left(x-\frac{5}{6}\right)^2\le\frac{49}{12}\)

Dấu "=" xảy ra khi x=5/6

b)\(D=-8x^2+4xy-y^2+3=3-\left(8x^2-4xy+y^2\right)=3-\left[\left(4x^2-4xy+y^2\right)+4x^2\right]\)

\(=3-\left[\left(2x-y\right)^2+4x^2\right]\le3\)

Dấu "=" xảy ra khi x=y=0

Bài 1:

a)\(3x^2+5x+2\)

\(=3\left(x+\frac{5}{6}\right)^2-\frac{1}{12}\ge-\frac{1}{12}\)

Dấu = khi \(x=-\frac{5}{6}\)

b)\(4x^2+y^2-2xy+7x-4y+10\)

tương tự có Min=\(\frac{21}{4}\Leftrightarrow x=-\frac{1}{2};y=\frac{3}{2}\)

Câu 2: ở đây Câu hỏi của Phạm Thùy Linh - Toán lớp 8 | Học trực tuyến

Câu 3:

\(a^{10}+b^{10}=a^{11}+b^{11}\)

\(\Rightarrow a^{11}-a^{10}+b^{11}-b^{10}=0\)

\(\Rightarrow a^{10}\left(a-1\right)+b^{10}\left(b-1\right)=0\left(1\right)\)

\(a^{10}\left(a-1\right)+b^{10}\left(b-1\right)>0\) không đúng với (1)

\(a^{10}\left(a-1\right)+b^{10}\left(b-1\right)< 0\) không đúng với (1)

Không mất tính tổng quát, giả sử \(a\ge1;b\le1\)

Ta có:

\(a^{10}\left(a-1\right)+b^{10}\left(b-1\right)=0\)

\(\Rightarrow a^{10}\left(a-1\right)=b^{10}\left(b-1\right)\left(2\right)\)

Lại có:

\(a^{11}+b^{11}=a^{12}+b^{12}\)

\(\Rightarrow a^{12}-a^{11}+b^{12}-b^{11}=0\)

\(\Rightarrow a^{11}\left(a-1\right)+b^{11}\left(b-1\right)=0\)

\(\Rightarrow a\cdot a^{10}\left(a-1\right)+b\cdot b^{10}\left(b-1\right)=0\)

\(\Rightarrow a\cdot a^{10}\left(a-1\right)-b\cdot b^{10}\left(b-1\right)=0\)

\(\Rightarrow a\cdot a^{10}\left(a-1\right)-b\cdot a^{10}\left(a-1\right)=0\)(theo (2))

\(\Rightarrow a^{100}\left(a-1\right)\left(a-b\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}a-1=0\\a-b=0\end{array}\right.\)(do a>0)

\(\Rightarrow a=b=1\Rightarrow A=1^{2015}+1^{2016}=2\)

bai dai dong qua

a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0

\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)

\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)

\(-5x-8=0\)

\(x=-\frac{8}{5}\)

Mai mik làm mấy bài kia sau

2/

b) ( cái bài này chịu)

c) (x+1)^3-(x-1)^3-6(x-1)^2=-10

(x+1-x+1)\(\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)\(-6\left(x^2-2x+1\right)=-10\)

\(2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6x^2+12x-6=-10\)

\(2\left(3x^2+1\right)-6x^2+12x-6=0\)

\(6x^2+2-6x^2+12x-6=-10\)

\(12x=-10+4\)

\(12x=-6=>x=-\frac{1}{2}\)

d) (5x-1)^2-(5x-4)(5x+4)=7

\(25x^2-10x+1-25x^2+16=7\)

-10x = 7 - 17

-10x = -10

x= 1

Câu còn lại bn làm tương tự

3/

a) 

Ta có: 

(a+b+c)^2=3(ab+bc+ca)

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 3ab + 3bc + 3ac

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 3ab - 3bc - 3ac = 0

a^2 + b^2 + c^2  - ac - bc - ab = 0

2a^2 + 2b^2 + 2c^2  - 2ac - 2bc - 2ab = 0

(a²-2ab+b²⁾+(a²-2ac+c²) + (b²-2bc +c²) = 0

(a-b)^2 + (a-c)^2 + (b-c)^2 =0

=> a=b=c