\(a,\left(x.\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x.\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ ---\\ b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{5}=\left(\dfrac{2}{\sqrt{5}}\right)^2=\left(-\dfrac{2}{\sqrt{5}}\right)^2 \\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\\x=-\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\end{matrix}\right.\\ Vậy:x=\pm\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\)

\(c,\left|3x-\dfrac{4}{5}\right|=\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}=\dfrac{11}{5}\\3x-\dfrac{4}{5}=-\dfrac{11}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=\dfrac{11}{5}+\dfrac{4}{5}=3\\3x=-\dfrac{11}{5}+\dfrac{4}{5}=-\dfrac{7}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{3}=1\\x=-\dfrac{7}{5}:3=-\dfrac{7}{15}\end{matrix}\right.\\ ---\\ d,\left|2x-2\right|=0\\ \Leftrightarrow2x-2=0\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\)

a: (x*1/2)^3=1/27

=>x*1/2=1/3

=>x=1/3:1/2=2/3

b: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{5}}{5}-\dfrac{1}{2}=\dfrac{4\sqrt{5}-5}{10}\\x=\dfrac{-4\sqrt{5}-5}{10}\end{matrix}\right.\)

c: =>3x-4/5=11/5 hoặc 3x-4/5=-11/5

=>3x=3 hoặc 3x=-7/5

=>x=-7/15 hoặc x=1

d: =>2x-2=0

=>2x=2

=>x=1

Bài 2:

1: =>5x+1=6/7 hoặc 5x+1=-6/7

=>5x=-1/7 hoặc 5x=-13/7

=>x=-1/35 hoặc x=-13/35

2: =>x-1=4

=>x=5

3: =>3x-1=3

=>3x=4

=>x=4/3

4: \(\Leftrightarrow\dfrac{5}{x+3}=\dfrac{-5}{6}+\dfrac{1}{2}=\dfrac{-5+3}{6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)

=>x+3=-15

=>x=-18

7: \(\Leftrightarrow2^{2x+1}+2^{2x+6}=264\)

=>2^2x+1*(1+2^5)=264

=>2^2x+1=8

=>2x+1=3

=>x=1

9: =>x^4=8x

=>x^4-8x=0

=>x=2

\(a,x-\dfrac{5}{7}=\dfrac{19}{21}\\ x=\dfrac{34}{21}\\ b,\dfrac{5}{3}-\left|x-\dfrac{1}{5}\right|=\dfrac{1}{3}\\ \left|x-\dfrac{1}{5}\right|=\dfrac{4}{3}\\ TH1:x-\dfrac{1}{5}=\dfrac{4}{3}\\ x=\dfrac{23}{15}\\ TH2:x-\dfrac{1}{5}=-\dfrac{4}{3}\\ x=-\dfrac{17}{15}\\ c,x-\dfrac{2}{5}=\dfrac{1}{4}\\ x=\dfrac{13}{20}\\ d,5\sqrt{x}-30=15\\ 5\sqrt{x}=45\\ \sqrt{x}=9\\ x=9^2=81\)

2. Tính:

a, \(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

=\(\left(\dfrac{-1}{20}+\dfrac{-1}{72}\right)+\left(\dfrac{-1}{30}+\dfrac{-1}{90}\right)+\left(\dfrac{-1}{42}+\dfrac{-1}{56}\right)\)

=\(\left(\dfrac{-18}{360}+\dfrac{-5}{360}\right)+\left(\dfrac{-3}{90}+\dfrac{-1}{90}\right)+\left(\dfrac{-4}{168}+\dfrac{-3}{168}\right)\)

=\(\dfrac{-23}{360}+\dfrac{-4}{90}+\dfrac{-7}{168}\)

=\(\dfrac{-23}{360}+\dfrac{-16}{360}+\dfrac{-15}{360}\)=\(\dfrac{-54}{360}=\dfrac{-3}{20}\)

b, \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

=\(\dfrac{5}{2}+\dfrac{4}{1}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{1}{15}+\dfrac{1}{15}.\dfrac{13}{4}\)

=\(\dfrac{5}{2}+\dfrac{1}{11}.\left(\dfrac{4}{1}+\dfrac{3}{2}\right)+\dfrac{1}{15}.\left(\dfrac{1}{2}+\dfrac{13}{4}\right)\)

=\(\dfrac{5}{2}+\dfrac{1}{11}.\dfrac{11}{2}+\dfrac{1}{15}.\dfrac{15}{4}\)

=\(\dfrac{5}{2}+\dfrac{1}{2}+\dfrac{1}{4}\)

=\(\dfrac{10}{4}+\dfrac{2}{4}+\dfrac{1}{4}\)

=\(\dfrac{13}{4}\)

3. Tìm x

a, \(\dfrac{x-5}{8}=\dfrac{18}{x-5}\)

\(\left(x-5\right).\left(x-5\right)=8.18\)

\(\left(x-5\right)^2=144\)

\(x-5=\sqrt{144}\)

\(x-5=12\)

\(x=12+5\)

\(x=17\)

b,\(\left(x-2\right)^{10}=\left(2-x\right)^8\)

\(x^{10}-2^{10}=x^8-2^8\)

\(x^{10}+x^8=2^{10}+2^8\)

\(\Rightarrow x=2\)

2b, Đặt giá trị biểu thức trên là B.

Ta có: \(B=\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

\(B=\dfrac{7}{7}\left(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\right)\)

\(B=7.\left(\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\right)\)

\(B=7.\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\right)\)

\(B=7.\left(\dfrac{1}{2}-\dfrac{1}{28}\right)\)

\(B=7.\left(\dfrac{14}{28}-\dfrac{1}{28}\right)=7.\dfrac{13}{28}=\dfrac{13}{4}\)

\(\dfrac{\left(13\dfrac{1}{4}-1\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)

\(=\dfrac{1\dfrac{25}{108}.230\dfrac{1}{25}+46\dfrac{3}{4}}{4\dfrac{16}{21}:\left(-1\dfrac{20}{21}\right)}=\dfrac{330\dfrac{1}{25}}{-2\dfrac{18}{41}}=-135,3164\)

Câu 2: 

(x-4/7)(x+1/2)>0

=>x-4/7>0 hoặc x+1/2<0

=>x>4/7 hoặc x<-1/2

Bài 1:

a: =>3x-3-4=0

=>3x=7

hay x=7/3

b: =>2x-2+3x+6=0

=>5x+4=0

hay x=-4/5

c: =>\(4x^2+4x-1=0\)

hay \(x\in\left\{\dfrac{-1+\sqrt{2}}{2};\dfrac{-1-\sqrt{2}}{2}\right\}\)

d: \(\Leftrightarrow3x-3+2x-4+6=0\)

=>5x+1=0

hay x=-1/5

Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ 

Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)

\(\left(-2;-1\right):+\)

\(\left(-1;1\right):-\)

\(\left(1;2\right):+\)

\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)

\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)

\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)

Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính 

2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)

\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)

\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)

3/ \(I=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx+\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

Xét \(A=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx\)

\(t=\sin x\Rightarrow dt=\cos x.dx\Rightarrow A=\int\limits^{\dfrac{\pi}{2}}_0e^t.dt=e^{\sin x}|^{\dfrac{\pi}{2}}_0\)

Xét \(B=\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

\(=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{1+\cos2x}{2}.dx=\dfrac{1}{2}.\int\limits^{\dfrac{\pi}{2}}_0dx+\dfrac{1}{2}\int\limits^{\dfrac{\pi}{2}}_0\cos2x.dx\)

\(=\dfrac{1}{2}x|^{\dfrac{\pi}{2}}_0+\dfrac{1}{2}.\dfrac{1}{2}\sin2x|^{\dfrac{\pi}{2}}_0\)

I=A+B=...