a/ \(x^2-2x=-1\)

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

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

\(\Leftrightarrow x-1=0\Rightarrow x=1\)

Vậy..............

b/ \(x^2+2x+1=0\)

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

\(\Leftrightarrow x+1=0\Rightarrow x=-1\)

Vậy.......

c/ \(4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)

\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)=3x^2\)

\(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)

\(\Leftrightarrow-4x=0\Rightarrow x=0\)

Vậy...................

d/ \(x\left(x-2017\right)-x^2\left(2017-x\right)=0\)

\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)

\(\Leftrightarrow x^3-2016x^2-2017x=0\)

\(\Leftrightarrow x^3+x^2-2017x^2-2017x=0\)

\(\Leftrightarrow x\left(x^2+x\right)-2017\left(x^2+x\right)=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x-2017\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\x-2017=0\Rightarrow x=2017\end{matrix}\right.\)

Vậy pt có 3 nghiệm là.....(tự ghi ra)

\(a,x^2-2x=-1\)

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

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

\(\Rightarrow x-2=0\Rightarrow x=2\)

\(b,x^2+2x+1=0\)

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

\(\Rightarrow x+2=0\Rightarrow x=-2\)

\(c,4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)

\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)-3x^2=0\) \(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)

\(\Leftrightarrow-4x=0\Rightarrow x=0\)

\(d,x\left(x-2017\right)-x^2\left(2017-x\right)=0\)

\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)

\(\Leftrightarrow x^3+x^2-2017x-2017=0\)

\(\Leftrightarrow x^2\left(x+1\right)-2017\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2017\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-2107=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x^2=2017\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\\left[{}\begin{matrix}x=\sqrt{2017}\\x=-\sqrt{2017}\end{matrix}\right.\end{matrix}\right.\)

Trần văn ổi ()

đù khó thế

tl j z mấy chế , k câu dc đâu :))

1) a) \(3x\left(x-\dfrac{2}{3}\right)=0\Leftrightarrow\left\{{}\begin{matrix}3x=0\\x-\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

vậy \(x=0;x=\dfrac{3}{2}\)

b) \(7\left(x-1\right)+2x\left(1-x\right)=0\Leftrightarrow7x-7+2x-2x^2=0\)

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

\(\Delta=9^2-4.\left(-2\right)\left(-7\right)=81-56=25>0\)

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(x_1=\dfrac{-9+5}{-4}=1\)

\(x_2=\dfrac{-9-5}{-4}=\dfrac{7}{2}\)

vậy \(x=1;x=\dfrac{7}{2}\)

câu 4 thế vào ; bấm máy tính

B1:

\(a,3x\left(x-\dfrac{2}{3}\right)=0\)

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

B4:

\(A=3x+8xy+3y\)

\(=3\left(x+y\right)+8xy\)

Mà \(x+y=\dfrac{4}{3}\) và \(xy=-2\)

\(\Rightarrow3.\dfrac{4}{3}+8.\left(-2\right)=4-16=-12\)

MK nghĩ là đề câu B sai ,theo mk :

\(B=x^5-5x^4+5x^3+5x^2+x+2017\)

Ta có : 5 = 4+ 1 = x+ 1

Thay 5 = x+1 vào B ,có :

= \(x^5-\left[\left(x+1\right)x^4\right]+\left(x+1\right)x^3+\left(x+1\right)x^2+x+2017\)

= \(x^5-\left(x^5+x^4\right)+x^4+x^3-x^3-x^2+x^2+x+2017\)

= \(x^5-x^5-x^4+x^4+x+2017\)

= x + 2017

= 4 + 2017

= 2021

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

cac ban oi ai xong truoc mk k cho nhe

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)