\(x^2\le4\)

\(\Leftrightarrow x^2\le2^2\)

\(\Leftrightarrow x=\left\{0;1;2;-1;-2\right\}\)

Thử lại : ta được kết quả đúng như trên

bạn ơi đề thiếu

\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)

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

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

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

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

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

__

`(x+2)(5-3x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

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

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

__

\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

Ko hiểu đề!

\(\left(3\frac{1}{2}+2x\right).2\frac{2}{3}=5\frac{1}{3}\)

<=>\(\left(\frac{7}{2}+2x\right).\frac{8}{3}=\frac{16}{3}\)

<=>\(\frac{28}{3}+\frac{16x}{3}=\frac{16}{3}\)

<=>\(\frac{16x}{3}=\frac{-2}{3}\)

<=>\(16x=-2\)

<=>\(x=\frac{-1}{8}\)

vậy \(x=\frac{-1}{8}\)

b,\(\left|2x+3\right|=5\)

xét x<0,ta co: \(\left|2x+3\right|=5\)<=> \(-2x+3=5\)<=>\(-2x=2\)<=>\(x=-1\)(loại)

xét x>0,ta co:\(\left|2x+3\right|=5\)<=>\(2x+3=5\)<=>\(2x=2\)<=>\(x=1\)

c,\(\frac{x-2}{4}=\frac{5+x}{3}\)

<=>\(\frac{3x-6}{12}=\frac{20+4x}{12}\)

=>\(3x-6=20+4x\)

<=>\(3x-6-20-4x=0\)

<=>\(-x-26=0\)

<=>\(-x=26\)

<=>\(x=-26\)

kl:.......

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

=> \(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\frac{x+349}{5}=0+1+1+1+1\)

=> \(\frac{x+2+327}{327}+\frac{x+3+326}{326}+\frac{x+4+325}{325}+\frac{x+5+324}{324}+\frac{x+329+20}{5}=4\)

=> \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}+\frac{20}{5}=4\)

=> \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)+4=4\)

=> \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Ta có : \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)

=> \(x+329=0\)

=> \(x=-329\)

Ko biết lm ns luôn đi,từ nãy tới h nhìn đi nhìn lại cái câu này ở mấy câu hỏi của bn rồi.

\(\left(x+\frac{1}{2}\right)^4=16\)

\(\Rightarrow\left[\begin{array}{nghiempt}\left(x+\frac{1}{2}\right)^4=2^4\\\left(x+\frac{1}{2}\right)^4=-2^4\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2\\x+\frac{1}{2}=-2\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=2-\frac{1}{2}\\x=-2-\frac{1}{2}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{5}{2}\end{array}\right.\)

             ( x + \(\frac{1}{2}\) )⁴ = 16

              Vì 2⁴ = 16 \(\Rightarrow\)x + \(\frac{1}{2}\) = 2

                                     x = 2 - \(\frac{1}{2}\)

                                      x = \(\frac{3}{2}\)

\(\left(x+\frac{1}{4}\right)^4=16\)

\(\Rightarrow\left[\begin{array}{nghiempt}\left(x+\frac{1}{4}\right)^4=2^4\\\left(x+\frac{1}{4}\right)^4=-2^4\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{4}=2\\x+\frac{1}{4}=-2\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=2-\frac{1}{4}\\x=-2-\frac{1}{4}\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{7}{4}\\x=-\frac{9}{4}\end{array}\right.\)

M có giá trị lớn nhất 

<=> (x + 1)² + 1 có giá trị nhỏ nhất

(x + 1)² lớn hơn hoặc bằng 0

=> (x + 1)² + 1 lớn hơn hoặc bằng 1

\(\Rightarrow\frac{7}{\left(x+1\right)^2+1}\le7\)

Vậy Max M = 7 khi x = 0

Ta có:

 .Để M đạt giá trị lớn nhất thì (x+1)² +1 phải nhỏ nhất.Vì (x+1)² > 0.

Vậy giá thị nhỏ nhất của (x+1)² +1=1.Vậy GTLN của M bằng 7 khi x = -1