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2 tháng 5 2017

a.

\(\left(x-1\right)\left(x+5\right)=0\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

b)

\(B\left(x\right)=5x^3-20x=0\)

\(\Leftrightarrow x\cdot\left(5x^2-20\right)=0\Rightarrow\orbr{\begin{cases}x=0\\5x^2-20=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\end{cases}}}\)\(\Leftrightarrow x\cdot\left(5x^2-20\right)\Rightarrow\orbr{\begin{cases}x=0\\5x^2-20=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\end{cases}}}\)

11 tháng 2 2022

\(A\left(x\right)=5x^2-5x+3=5\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0,\forall x\)

⇒ pt vô nghiệm

\(B\left(x\right)=4x^2-3x+7=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{103}{16}>0,\forall x\)

⇒ pt vô nghiệm

\(C\left(x\right)=5x^2-11x+6=\left(5x^2-5x\right)-\left(6x-6\right)\)

\(=5x\left(x-1\right)-6\left(x-1\right)=\left(5x-6\right)\left(x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=1\end{matrix}\right.\)

Vậy ...

11 tháng 2 2022

a, Ta có : 

\(A\left(x\right)=5x^2-5x+1+2=0\Leftrightarrow5x^2-6x+3=0\)

\(\Leftrightarrow5\left(x^2-\dfrac{2.3}{5}+\dfrac{9}{25}-\dfrac{9}{25}\right)+3=0\Leftrightarrow5\left(x-\dfrac{3}{5}\right)^2+\dfrac{6}{5}=0\)( vô lí )

vậy đa thức ko có nghiệm 

b, \(B\left(x\right)=4x^2-3x+7=0\Leftrightarrow4\left(x^2-\dfrac{2.3}{8}+\dfrac{9}{64}-\dfrac{9}{64}\right)+7=0\)

\(\Leftrightarrow4\left(x-\dfrac{3}{8}\right)^2+\dfrac{103}{64}=0\)( vô lí ) 

Vậy đa thức ko có nghiệm 

c, \(C\left(x\right)=5x^2-11x+6=0\Leftrightarrow5x^2-6x-5x+6=0\)

\(\Leftrightarrow5x\left(x-1\right)-6\left(x-1\right)=0\Leftrightarrow\left(5x-6\right)\left(x-1\right)=0\Leftrightarrow x=\dfrac{6}{5};x=1\)

Bài 1:

a: ĐKXĐ: \(x+4\ne0\)

=>\(x\ne-4\)

b: ĐKXĐ: \(2x-1\ne0\)

=>\(2x\ne1\)

=>\(x\ne\dfrac{1}{2}\)

c: ĐKXĐ: \(x\left(y-3\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne0\\y-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\)

d: ĐKXĐ: \(x^2-4y^2\ne0\)

=>\(\left(x-2y\right)\left(x+2y\right)\ne0\)

=>\(x\ne\pm2y\)

e: ĐKXĐ: \(\left(5-x\right)\left(y+2\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne5\\y\ne-2\end{matrix}\right.\)

 Bài 2:

a: \(\dfrac{-12x^3y^2}{-20x^2y^2}=\dfrac{12x^3y^2}{20x^2y^2}=\dfrac{12x^3y^2:4x^2y^2}{20x^2y^2:4x^2y^2}=\dfrac{3x}{5}\)

b: \(\dfrac{x^2+xy-x-y}{x^2-xy-x+y}\)

\(=\dfrac{\left(x^2+xy\right)-\left(x+y\right)}{\left(x^2-xy\right)-\left(x-y\right)}\)

\(=\dfrac{x\left(x+y\right)-\left(x+y\right)}{x\left(x-y\right)-\left(x-y\right)}=\dfrac{\left(x+y\right)\left(x-1\right)}{\left(x-y\right)\left(x-1\right)}\)

\(=\dfrac{x+y}{x-y}\)

c: \(\dfrac{7x^2-7xy}{y^2-x^2}\)

\(=\dfrac{7x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}\)

\(=\dfrac{-7x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{-7x}{x+y}\)
d: \(\dfrac{7x^2+14x+7}{3x^2+3x}\)

\(=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

e: \(\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}\)

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

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

g: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)

\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{x+4}{x+2}\)

 

20 tháng 10 2021

\(a,A=\left(x^2+5x+\dfrac{25}{4}\right)+\dfrac{7}{4}=\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\\ A_{min}=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{5}{2}\\ b,B=x^2-6x+9-9=\left(x-3\right)^2-9\ge9\\ B_{min}=-9\Leftrightarrow x=3\)

30 tháng 3 2018

b) x2+5x-6 =0

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

\(\Leftrightarrow x\left(x+6\right)-\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)

Vậy S = {-6;1}

c) x2-4x+3=0

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

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

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

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

Vậy S = {3;1}

d) 2x2+5x+3=0

\(\Leftrightarrow2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)

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

Vậy S = {-1;\(\dfrac{-3}{2}\)}

30 tháng 3 2018

bài 2

\(\left(x-1\right)^2+\left(x+5\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\) (vô lí)

Vậy pt vô nghiệm

20 tháng 6 2023

\(a,\) Đặt \(P\left(x\right)=0\Rightarrow x^2+3x=0\Rightarrow x\left(x+3\right)=0\)

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

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

Vậy nghiệm của P(x) là \(x=0;x=-3\)

\(a,\) Đặt \(P\left(x\right)=0\Rightarrow3x^2-15=0\Rightarrow3\left(x^2-5\right)=0\)

\(\Rightarrow x=\pm\sqrt{5}\)

Vậy nghiệm của P(x) là \(x=\pm\sqrt{5}\)

 

20 tháng 6 2023

Đề yc tìm nghiệm em ơi

4 tháng 5 2023

\(Câu8\)

\(a,A=\dfrac{1}{2}x^3\times\dfrac{8}{5}x^2=\left(\dfrac{1}{2}\times\dfrac{8}{5}\right)x^{3+2}=\dfrac{4}{5}x^5\)

b, \(P\left(0\right)=0^2-5.0+6=6\\ P\left(2\right)=2^2-5.2+6=0\)

Câu 9

\(a,A\left(x\right)+B\left(x\right)=5x^3+x^2-3x+5+5x^3+x^2+2x-3\\ =\left(5x^3+5x^3\right)+\left(x^2+x^2\right)+\left(-3x+2x\right)+\left(5-3\right)\\ =10x^3+2x^2-x+2\)

\(b,H\left(x\right)=A\left(x\right)-B\left(x\right)=5x^3+x^2-3x+5-\left(5x^3+x^2+2x-3\right)\\ =5x^3+x^2-3x+5-5x^3-x^2-2x+3\\ =\left(5x^3-5x^3\right)+\left(x^2-x^2\right) +\left(-3x-2x\right)+\left(5+3\right)\\ =-5x+8\)

\(H\left(x\right)=0\\ \Rightarrow-5x+8=0\\ \Rightarrow x=\dfrac{8}{5}\)

vậy nghiệm của đa thức là \(x=\dfrac{8}{5}\)

2 tháng 8 2017

a) Thay x = 4 vào biểu thức A :

A = 45 - 5.44+ 5.43 - 5.42 + 5.4 -1

= 3

b) Thay x = 21 vào B :

B = 216 - 20.215 - 20.214 -20.213 - 20.212 - 20.21+3

=24

a: A(x)=0

=>5x-7=0

=>x=7/5

b: P(x)=0

=>x-1=0 hoặc x+3=0

=>x=1 hoặc x=-3

c: Q(x)=0

=>(2/3x-1)=0 hoặc x+3/5=0

=>x=-3/5 hoặc x=3/2

14 tháng 9 2021

a) \(4x+9=0\Leftrightarrow4x=-9\Leftrightarrow x=-\dfrac{9}{4}\)

b) \(-5x+6=0\Leftrightarrow5x=6\Leftrightarrow x=\dfrac{6}{5}\)

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

d) \(x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

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

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

g) \(\left(x-4\right)\left(x^2+1\right)=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)( do \(x^2+1\ge1>0\))

h) \(3x^2-4x=0\Leftrightarrow x\left(3x-4\right)=0\Leftrightarrow\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

i) \(x^2+9=0\Leftrightarrow x^2=-9\)( vô lý do \(x^2\ge0>-9\))

Vậy \(x\in\left\{\varnothing\right\}\)