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28 tháng 7 2017

(x- 1)(x- 3)(x- 5)(x- 7)- 20

= (x- 1)(x- 7)(x- 3)(x- 5)- 20

= (x2- 8x+ 7)(x2- 8x+ 15)- 20

Đătk x2- 8x+ 11= t

Ta có (t- 4)(t+ 4) -20

= t2- 16- 20

= t2- 36= (t- 6)(t+ 6)

Thay t= x2- 8x+ 11

=> (x2- 8x+ 11- 6)(x2- 8x +11+ 6)

= (x2- 8x+ 5)(x2- 8x+ 17)

= (x2- 8x+ 16- 11)(x2- 8x+ 17)

= ((x- 4)2- 11)(x2- 8x+ 17)

\(=\left(x-4-\sqrt{11}\right)\left(x-4+\sqrt{11}\right)\left(x^2-8x+17\right)\)

28 tháng 7 2017

Đặt đa thức trên là A
<=>A=(x-1)(x-7)(x-3)(x-5)-20
<=>A=(x2-8x-7)(x2-8x+15)-20
Đặt t=x2-8x-7, ta có:

A=t(t+22)-20
   =t2+22t-20
   =t2+22t+121-141
   =(t+11)2-141
thayt=x2-8x- 7vào
 

12 tháng 10 2017

Bài 1

1.(x-3)(x+2)-x(x-7)=15

\(\Leftrightarrow x^2+2x-3x-6-x^2+7x=15\)

\(\Leftrightarrow-6+6x=15\)

\(\Leftrightarrow6x=15+6\) =21

\(\Rightarrow x=\dfrac{21}{6}=3,5\)

2.(x-5)(x+5)+x(3-x)=20

\(\Leftrightarrow x^2-25+3x-x^2=20\)

\(\Leftrightarrow-25+3x=20\)

\(\Leftrightarrow3x=20+25=45\)

\(\Rightarrow x=\dfrac{45}{3}=15\)

3.(x-7)2-x(2+x)=-7

\(\Leftrightarrow x^2-14x+49-2x-x^2=-7\)

\(\Leftrightarrow-16x+49=-7\)

\(\Leftrightarrow-16x=-7-49=-56\)

\(\Rightarrow x=\dfrac{-56}{-16}=\dfrac{7}{2}=3,5\)

12 tháng 10 2017

Tiếp bài 1

4.(x-4)2-(x+4)(x-4)=-16

\(\Leftrightarrow x^2-8x+16-x^2-16=-16\)

\(\Leftrightarrow-8x=-16\)

\(\Rightarrow x=\dfrac{-16}{-8}=2\)

5.(x-5)(x+5)-x(2-3x)=4x2-7

\(\Leftrightarrow x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-4x^2-2x=-7+25\)

\(\Leftrightarrow-2x=18\)

\(\Rightarrow x=\dfrac{18}{-2}=-9\)

3 tháng 3 2022

\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)=20\\ \Leftrightarrow\left[\left(x-1\right)\left(x-7\right)\right]\left[\left(x-3\right)\left(x-5\right)\right]-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)\left[\left(x^2-8x+7\right)+8\right]-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)^2+8\left(x^2-8x+7\right)-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)^2-2\left(x^2-8x+7\right)+10\left(x^2-8x+7\right)-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)\left(x^2-8x+7-2\right)+10\left(x^2-8x+7-2\right)=0\)

\(\Leftrightarrow\left(x^2-8x+7+10\right)\left(x^2-8x+7-2\right)=0\\ \Leftrightarrow\left(x^2-8x+17\right)\left(x^2-8x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-8x+16+1=0\\x^2-8x+16-11=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-4\right)^2+1=0\left(vô.lí\right)\\\left(x-4\right)^2-11=0\end{matrix}\right.\\ \Leftrightarrow\left(x-4-\sqrt{11}\right)\left(x-4+\sqrt{11}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{11}\\x=4-\sqrt{11}\end{matrix}\right.\)

3 tháng 3 2022

\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)

\(=\left[\left(x-1\right)\left(x-7\right)\right]\left[\left(x-3\right)\left(x-5\right)\right]-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(x^2-8x+7=t\),ta có :

\(t\left(t+8\right)-20\)

\(=t^2+8t-20\)

\(=\left(t^2+8t+16\right)-16-20\)

\(=\left(t+4\right)^2-36\)

\(=\left(t+4\right)^2-6^2\)

\(=\left(t+4-6\right)\left(t+4+6\right)\)

\(=\left(t-2\right)\left(t+10\right)\)

\(=\left(x^2-8x+7-2\right)\left(x^2-8x+7+10\right)\)

\(=\left(x^2-8x+5\right)\left(x^2-8x+17\right)\)

5 tháng 12 2017

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

a: \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)

=>1+3x-6=3-x

=>3x-5=3-x

=>4x=8

hay x=2(loại)

b: \(\Leftrightarrow8-x-8\left(x-7\right)=-26\)

=>8-x-8x+56=-26

=>-9x+64=-26

=>-9x=-90

hay x=10(nhận)

c: \(\dfrac{1}{x-2}+\dfrac{1}{x-3}=\dfrac{2}{x-1}\)

\(\Leftrightarrow\dfrac{x-3+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2}{x-1}\)

\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=2\left(x^2-5x+6\right)\)

\(\Leftrightarrow2x^2-5x-2x+5=2x^2-10x+12\)

=>-7x+10x=12-5

=>3x=7

hay x=7/3(nhận)

 

undefined

3: \(-\dfrac{1}{2}+\left|\dfrac{1}{3}x+\dfrac{2}{5}\right|=\dfrac{5}{7}\cdot\dfrac{3}{11}+\dfrac{8}{11}\cdot\dfrac{5}{7}=\dfrac{5}{7}\)

=>|1/3x+2/5|=5/7+1/2=17/14

=>1/3x+2/5=17/14 hoặc 1/3x+2/5=-17/14

=>x=171/70 hoặc x=-339/70

28 tháng 9 2019

1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x=t\)

\(\Rightarrow BT=\left(t+10\right)\left(t+12\right)-24\)

\(=t^2+22x+96=\left(t+11\right)^2-25\ge-25\)

Vậy GTNN của bt là - 25\(\Leftrightarrow x^2+7x+11=0\)

\(\Delta=7^2-4.11=5\)

\(\orbr{\begin{cases}x_1=\frac{-22+\sqrt{5}}{2}\\x_2=\frac{-22-\sqrt{5}}{2}\end{cases}}\)

28 tháng 9 2019

2) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)

\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(x^2-8x=t\)

\(\RightarrowĐT=\left(t+7\right)\left(t+15\right)-20\)

\(=t^2+22t+85=\left(t+11\right)^2-36\ge-36\)

Vậy GTNN của bt là - 36\(\Leftrightarrow x^2-8x+11=0\)

\(\Delta=\left(-8\right)^2-4.11=20\)

\(\orbr{\begin{cases}x_1=\frac{-22-\sqrt{20}}{2}\\x_2=\frac{-22+\sqrt{20}}{2}\end{cases}}\)

11 tháng 3 2023

`a,4x-10=0   `

`<=> 4x=10`

`<=>x=10/4`

`<=>x=5/2`

`b, 7-3x=9-x     `

`<=>-3x+x=9-7`

`<=>-2x=2`

`<=>x=-1`

`c, 2x-(3-5x) = 4(x+3)`

`<=>2x-3+5x=4x+12`

`<=>2x+5x-4x=12+3`

`<=>3x=15`

`<=>x=5`

`d, 5-(6-x)=4(3-2x)     `

`<=>5-6+x=12-8x`

`<=>x+8x=12-5+6`

`<=>9x=13`

`<=>x=13/9`

`e, 4(x+3)=-7x+17   `

`<=>4x+12=-7x+17`

`<=>4x+7x=17-12`

`<=>11x=5`

`<=>x=5/11`   

`f, 5(x-3) - 4=2(x-1)+7`

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`g, 5(x-3)-4=2(x-1)+7       `

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`h,4(3x-2)-3(x-4)=7x+20`

`<=>12x-8-3x+12=7x+20`

`<=>12x-3x-7x=20+8+12`

`<=>2x=40`

`<=>x=20`