a) \(\left|x-5\right|-\left|x-7\right|\le\left|x-5-x+7\right|=2\)

Dấu "=" xảy ra: 

`(x-5)(x-7)<=0<=>5<=x<=7` 

b) \(\left|3x-5\right|-\left|7-3x\right|=\left|3x-5\right|-\left|3x-7\right|\le\left|3x-5-3x+7\right|=2\)

Dấu "=" xảy ra: 

`(3x-5)(3x-7)<=0<=>5/3<=x<=7/3` 

c) \(\left|1-x\right|-\left|2-x\right|\le\left|1-x-2+x\right|=1\)

Dấu "=" xảy ra: 

`(1-x)(2-x)<=0<=>(x-1)(x-2)<=0<=>1<=x<=2`

A=104-100+96-92+88-84+...-12+8

=(104-100)+(96-92)+...+(16-12)+8

=4+4+...+4+8

\(=4\cdot12+8=48+8=56\)

a: ĐKXĐ: x>=1/2

\(\sqrt{2x-1}=5\)

=>\(2x-1=5^2=25\)

=>2x=26

=>x=13(nhận)

b: ĐKXĐ: \(x>=-\dfrac{2}{3}\)

\(\sqrt{3x+2}=\dfrac{1}{4}\)

=>\(3x+2=\left(\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

=>\(3x=\dfrac{1}{16}-2=\dfrac{1}{16}-\dfrac{32}{16}=-\dfrac{31}{16}\)

=>\(x=-\dfrac{31}{48}\left(nhận\right)\)

c: \(\sqrt{x^2+\dfrac{1}{4}}=\sqrt{\dfrac{49}{81}}\)

=>\(x^2+\dfrac{1}{4}=\dfrac{49}{81}\)

=>\(x^2=\dfrac{49}{81}-\dfrac{1}{4}=\dfrac{115}{324}\)

=>\(x=\pm\dfrac{\sqrt{115}}{18}\)

\(1\dfrac{1}{5}:\left\{\dfrac{5}{8}+\left[\dfrac{5}{3}-\left(-\dfrac{1}{4}\right)\right]\cdot\dfrac{9}{2^2}\right\}\)

\(=\dfrac{6}{5}:\left\{\dfrac{5}{8}+\left(\dfrac{5}{3}+\dfrac{1}{4}\right)\cdot\dfrac{9}{4}\right\}\)

\(=\dfrac{6}{5}:\left\{\dfrac{5}{8}+\dfrac{23}{12}\cdot\dfrac{9}{4}\right\}\)

\(=\dfrac{6}{5}:\left\{\dfrac{5}{8}+\dfrac{23\cdot3}{16}\right\}=\dfrac{6}{5}:\left(\dfrac{10}{16}+\dfrac{69}{16}\right)\)

\(=\dfrac{6}{5}\cdot\dfrac{16}{79}=\dfrac{96}{395}\)

Thể tích nước trong thùng ban đầu là:

\(V_1=x\cdot a\cdot b\left(dm^3\right)\)

Diện tích đáy trong thùng sau khi nghiêng là:

\(S_{đáy}=\dfrac{1}{2}\cdot\dfrac{3}{4}a\cdot8=3a\left(dm^2\right)\)

Thể tích nước sau khi nghiêng thùng là: \(V_2=3a\cdot b\left(dm^3\right)\)

Vì thể tích nước trước và sau khi nghiêng thùng đều không thay đổi nên \(x\cdot a\cdot b=3\cdot a\cdot b\)

=>x=3

Ta có; ΔABC=ΔDEF

=>AB=DE; BC=EF; AC=DF; \(\widehat{BAC}=\widehat{EDF};\widehat{ABC}=\widehat{DEF};\widehat{ACB}=\widehat{DFE}\)

Xét ΔBAM và ΔEDN có

AB=DE

\(\widehat{ABM}=\widehat{DEN}\)

BM=EN

Do đó: ΔBAM=ΔEDN

=>AM=DN và \(\widehat{BAM}=\widehat{EDN}\)

a: \(\widehat{MON}+\widehat{O_1}+45^0=180^0\)

=>\(\widehat{O_1}=180^0-90^0-45^0=45^0\)

Ta có: \(\widehat{O_1}=\widehat{MNO}\left(=45^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên OB//AM

b: Ta có: OB//AM

MA\(\perp\)AB

Do đó: OB\(\perp\)BA

\(A=\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)

\(299A=1+\dfrac{1}{2}+...+\dfrac{1}{101}-\left(\dfrac{1}{300}+\dfrac{1}{301}+...+\dfrac{1}{400}\right)\)

Thêm bớt \(\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{299}\) ta được:

\(299A=1+\dfrac{1}{2}+...+\dfrac{1}{101}+\left(\dfrac{1}{102}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{300}+...+\dfrac{1}{400}\right)\)

\(299A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{400}\right)\)

\(101B=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+\dfrac{1}{3}-\dfrac{1}{104}+....+\dfrac{1}{299}-\dfrac{1}{400}\)

\(101B=\left(1+\dfrac{1}{2}+...+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{400}\right)\)

\(\Rightarrow299A=101B\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{101}{299}\)