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16 tháng 10 2016

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

b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)

d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)

e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)

g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)

h)Tương tự các câu trên

i) x = 0

k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)

l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)

=> x + 1 = 3 => x = 2

16 tháng 10 2016

x.(x+1)=0

suy ra x=0 hoac x+1=0

                               x=0-1

                              x=-1

vay x=0 hoac  x=-1

mấy câu sau cũng làm tương tự

19 tháng 6 2016

a) \(\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5=\left(\frac{3}{5}\right)^{15}:\left(\left(\frac{3}{5}\right)^2\right)^5=\left(\frac{3}{5}\right)^{15}:\left(\frac{3}{5}\right)^{10}=\left(\frac{3}{5}\right)^5\)

b) \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=4+\frac{1}{27}=4\frac{1}{27}\)

c) \(2^3+3.\left(\frac{1}{2}\right)^0+\left(-2\right)^2:\frac{1}{2}.8=8+3.1+4:\frac{1}{2}.8=8+3+64=75\)

d) \(\left(-1\right)^{-1}-\left(-\frac{3}{5}\right)^0+\left(\frac{1}{2}\right)^{2:2}=-1-1+\left(\frac{1}{2}\right)^1=-2+\frac{1}{2}=-\frac{3}{2}\)

19 tháng 6 2016

bạn giải thích cho mình phần a nhéhaha

a: \(\Leftrightarrow25\left(x+1\right)^4-25\left(x+1\right)^2-\left(x+1\right)^2+1=0\)

\(\Leftrightarrow\left[\left(x+1\right)^2-1\right]\left[25\left(x+1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x+2\right)\cdot x\cdot\left(5x+4\right)\left(5x+6\right)=0\)

hay \(x\in\left\{0;-2;-\dfrac{4}{5};-\dfrac{6}{5}\right\}\)

b: \(x^2+x-1=0\)

\(\text{Δ}=1^2-4\cdot1\cdot\left(-1\right)=5\)

Do đó: PT có 2 nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{5}}{2}\\x_2=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)

d: \(\Leftrightarrow4x^2-4x+1-5\left(2x-1\right)-6=0\)

\(\Leftrightarrow\left(2x-1\right)^2-5\left(2x-1\right)-6=0\)

=>(2x-1-6)(2x-1+1)=0

=>(2x-7)2x=0

=>x=0 hoặc x=7/2

25 tháng 7 2018

a, \(\left(x-3\right)\left(2x+5\right)>0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\2x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\2x+5< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< -\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -\dfrac{5}{2}\end{matrix}\right.\)

b,\(\left(1-4x\right)\left(x-2\right)< 0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-4x>0\\x-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-4x< 0\\x-2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{1}{4}\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{1}{4}\\x>2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 2\\x>2\end{matrix}\right.\)

25 tháng 7 2018

c, \(\dfrac{-3}{x+2}< 0\Leftrightarrow x+2>0\Leftrightarrow x>-2\)

29 tháng 8 2017

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) vậy \(x=3;x=1\)

c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=\sqrt[3]{-8}\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)

d) \(\left(x+2\right)^2+1=0\Leftrightarrow\left(x+2\right)^2=-1\) (vô lí)

vậy phương trình vô nghiệm

29 tháng 8 2017

a) (x-1)2 = 0

<=> x-1 = 0

<=> x = 1

b) (x-2)2 - 1 = 0

<=> (x-2)2 = 1

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

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

c) (2x-1)3 = -8

<=> (2x-1)3 = -23

<=> 2x - 1 = -2

<=> 2x = -1

<=> x = \(-\dfrac{1}{2}\)

d) (x+2)2 + 1 = 0

<=> (x+2)2 = -1

<=> x+2 = -1

<=> x = -3

5 tháng 1 2020

2.

a) \(3.\left(x-1\right)-2.\left|x+3\right|\)

TH1: \(x\ge-3.\)

\(3.\left(x-1\right)-2.\left|x+3\right|\)

\(=3x-3-2.\left(x+3\right)\)

\(=3x-3-\left(2x+6\right)\)

\(=3x-3-2x-6\)

\(=x-9.\)

TH2: \(x< -3.\)

\(3.\left(x-1\right)-2.\left|x+3\right|\)

\(=3.\left(x-1\right)-2.\left[-\left(x+3\right)\right]\)

\(=3x-3-2.\left(-x-3\right)\)

\(=3x-3-\left(-2x-6\right)\)

\(=3x-3+2x+6\)

\(=5x+3.\)

Chúc bạn học tốt!

5 tháng 1 2020

Bạn ơi phần a là như này đúng không ạ :

TH1 : \(x+3\ge0\Leftrightarrow x\ge-3\)

6 tháng 7 2017

b) \(\left|4-7x\right|-\dfrac{3}{2}:5=\left|-1\dfrac{1}{3}\right|\)

\(\left|4-7x\right|-\dfrac{3}{10}=\dfrac{4}{3}\)

\(\left|4-7x\right|=\dfrac{49}{30}\) (*)

+) Nếu 4 - 7x \(\ge\) 0 \(\Rightarrow x\le\dfrac{4}{7}\)

PT (*) \(\Leftrightarrow4-7x=\dfrac{49}{30}\)

\(-7x=-\dfrac{71}{30}\)

x = \(\dfrac{71}{210}\) (t/m)

+) Nếu \(4-7x< 0\Rightarrow x>\dfrac{4}{7}\)

Pt (*) \(\Leftrightarrow-4+7x=\dfrac{49}{30}\)

x = \(\dfrac{169}{210}\) (t/m)

Vậy x=\(\dfrac{71}{210}\) hoặc x = \(\dfrac{169}{210}\)

8 tháng 10 2017

*) \(4x^2-1=0\)

\(\Rightarrow4x^2=1\Rightarrow x^2=\dfrac{1}{4}\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

*) \(2x^2+0,82=1\)

\(\Rightarrow2x^2=1-0,82=\dfrac{9}{50}\)

\(\Rightarrow x^2=\dfrac{9}{100}\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{3}{10}\end{matrix}\right.\)

*) \(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{12}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

8 tháng 10 2017

Giải:

a) \(4x^2-1=0\)

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

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

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

Vậy ...

b) \(2x^2+0,82=1\)

\(\Leftrightarrow2x^2=0,18\)

\(\Leftrightarrow x^2=0,09\)

\(\Leftrightarrow x=\pm0,3\)

Vậy ...

c) \(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{12}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...

Chúc bạn học tốt!

25 tháng 8 2020

Bài làm:

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

=> \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=0\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=0\) (1)

Mà \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\), cách CM như sau:

\(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)

Tương tự: \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\) ; \(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\)

Cộng vế 3 BĐT trên lại ta sẽ được: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

Thay vào (1) ta được:

\(0=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\ge3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\le0\)

Dấu "=" xảy ra khi: \(a=b=c\)

24 tháng 6 2017

a) Ta có :

\(0,\left(27\right)+0,\left(72\right)==\dfrac{27}{99}+\dfrac{72}{99}=\dfrac{99}{99}=1\)

\(\Rightarrow0,\left(27\right)+0,\left(72\right)=1\rightarrowđpcm\)

b) Ta có :

\(0,\left(22\right).\dfrac{9}{2}=\dfrac{2}{9}.\dfrac{9}{2}=\dfrac{18}{18}=1\)

\(\Rightarrow0,22.\dfrac{9}{2}=1\rightarrowđpcm\)

c) Ta có :

\(\left[0,\left(11\right).9\right]^{2003}=\left[\dfrac{1}{9}.9\right]^{2003}=\left[\dfrac{9}{9}\right]^{2003}=1^{2003}=1\)

\(\Rightarrow\left[0,\left(11\right).9\right]^{2003}=1\rightarrowđpcm\)

24 tháng 6 2017

a) \(0,\left(27\right)+0,\left(72\right)=0,\left(99\right)=1\)

b) \(0,\left(22\right)\cdot\dfrac{9}{2}=\dfrac{2}{9}\cdot\dfrac{9}{2}=1\)

c) \(\left[0,\left(11\right)\cdot9\right]^{2003}=\left(\dfrac{1}{9}\cdot9\right)^{2003}=1^{2003}=1\)