Bn vào Link này tham khảo nhé bn 

Rút gọn biểu thức: √[x + 4√(x - 4)] + √[x - 4√(x - 4)] - Toán học Lớp 9 - Bài tập Toán học Lớp 9 - Giải bài tập Toán học Lớp 9 | Lazi.vn - Cộng đồng Tri thức & Giáo dục

# MissyGirl #

Giải giùm mình ạ, đừng đưa link nha😅

\(A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(=\sqrt{\sqrt{\left(x-4\right)^2}+2.2.\sqrt{x-4}+2^2}+\sqrt{\sqrt{\left(x-4\right)^2}-2.2.\sqrt{x-4}+2^2}\)

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

\(=\sqrt{x-4}+2+|\sqrt{x-4}-2|\)

Xet \(\sqrt{x-4}-2\ge0\)

\(\Rightarrow A=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)

Xet \(\sqrt{x-4}-2< 0\)

\(\Rightarrow A=\sqrt{x-4}+2-\sqrt{x-4}+2=4\) 

`(15-x)+(x-12)=7-(-5+x)`

`=>15-x+x-12=7+5-x`

`=>3=12-x`

`=>x=12-3`

`=>x=9`

Vậy `x=9`

(15-x)+(x-12) = 7-(-5+x)

<=>15-x+x-12=7+5-x

<=>3=12-x

<=>x=12-3=9

(15-x)+(x-12) = 7-(-5+x)

<=> 15-x +x -12 = 7 +5 -x 

<=> x = 12 +12 -15

<=> x =9

B=\(\left[\left(\frac{1}{3}+\frac{1}{4}\right)x\frac{12}{19}+\frac{12}{19}\right]:\frac{4}{5}-\frac{1}{4}+2012\)

B=\(\left(\frac{7}{12}x\frac{12}{19}+\frac{12}{19}\right):\frac{4}{5}-\frac{1}{4}+2012\)

B=\(\left(\frac{7}{19}+\frac{12}{19}\right):\frac{4}{5}-\frac{1}{4}+2012\)

B=\(\frac{5}{4}-\frac{1}{4}+2012\)

B=1+2012

B=2013

\(B=[\left(\frac{1}{3}+\frac{1}{4}\right)\times\frac{12}{19}+\frac{12}{19}]:\frac{4}{5}-\frac{1}{4}+2012\)

\(B=[\frac{7}{12}\times\frac{12}{19}+\frac{12}{19}]:\frac{4}{5}-\frac{1}{4}+2012\)

\(B=[\frac{7}{19}+\frac{12}{19}]:\frac{4}{5}-\frac{1}{4}+2012\)

\(B=1:\frac{4}{5}-\frac{1}{4}+2012\)

\(B=\frac{5}{4}-\frac{1}{4}+2012\)

\(B=1+2012\)

\(B=2013\)

\(B=\left[\left(\frac{1}{3}+\frac{1}{4}\right)\times\frac{12}{19}+\frac{12}{19}\right]:\frac{4}{5}-\frac{1}{4}+2012\)

\(B=\left[\frac{7}{12}\times\frac{12}{19}+\frac{12}{19}\right]:\frac{4}{5}-\frac{1}{4}+2012\)

\(B=\left[\left(\frac{7}{12}+1\right)\times\frac{12}{19}\right]:\frac{4}{5}-\frac{1}{4}+2012\)

\(B=\left[\frac{19}{12}\times\frac{12}{19}\right]:\frac{4}{5}-\frac{1}{4}+2012\)

\(B=1:\frac{4}{5}-\frac{1}{4}+2012\)

\(B=\frac{5}{4}-\frac{1}{4}+2012\)

\(B=1+2012\)

\(B=2013\)

Vậy B = 2013

_Chúc bạn học tốt_

\(a,\left(\dfrac{1}{x-1}-\dfrac{x}{x-1^2}.\dfrac{x^2+1+x}{x+1}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1}{x-1}-\dfrac{x\left(x^2+1+x\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^3+x+x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\)

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

b,

thay x=\(\dfrac{1}{2}\) vào bt M ta được:

\(1-\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2=\dfrac{5}{8}\)

 A = x + | x |

có ; \(\left|x\right|\ge0\forall x\)

=> \(x+\left|x\right|\ge x\forall x\)

dấu ''='' xảy ra <=> x =0

vậy gtnn của A là x tại x=0

b) ta có : \(\left|x-3\right|\ge0\forall x\in Z\)

dấu ''='' xảy ra <=> x-3=0

=>  x=3

vậy gtnn  của bt B là 0 tại x=3

c) | x - 2 | + | x - 4 |

\(C=\left|x-2\right|+\left|x-4\right|\ge\left|x-2\right|+\left|4-x\right|\ge\left|x-2+4-x\right|\ge2\)

dấu ''='' xảy ra <=> \(\left(x-2\right)\left(4-x\right)\ge0\)

\(\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=4\end{cases}}\)

vậy gtnn của bt C là 2 tại x ={2;4}

\(\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\frac{x^3\left(x-1\right)-\left(x-1\right)}{x^4+x^3+x^2+2x^2+2x+2}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+2\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x^2+2\right)}\)

a)

\(\dfrac{6}{13}\cdot\dfrac{8}{7}\cdot\dfrac{-26}{3}\cdot\dfrac{-7}{8}\)

\(=\dfrac{6}{13}\cdot\dfrac{-26}{3}\cdot\dfrac{8}{7}\cdot\dfrac{-7}{8}\)

\(=-4\cdot\left(-1\right)\\ =4\)

b)

\(\dfrac{6}{11}+\dfrac{11}{3}\cdot\dfrac{3}{22}\)

\(=\dfrac{6}{11}+\dfrac{1}{2}\\ =\dfrac{12}{22}+\dfrac{11}{22}\\ =\dfrac{23}{22}\)

a)=  (6/13*-26/3)* (8/7*-7/8)

= -4*-1=4

b=6/11+1/2=(12+11)/22=23/22

a. \(\dfrac{6.8.\left(-26\right).\left(-7\right)}{13.7.3.8}=\dfrac{2.3.8.\left(-13\right).2.\left(-7\right)}{13.7.3.8}=\dfrac{4}{1}=4\)

\(b.\dfrac{6}{11}+\dfrac{11.3}{3.11.2}\)

\(=\dfrac{6}{11}+\dfrac{1}{2}=\dfrac{23}{22}\)