Ta có: \(\left|6x-2\right|\cdot\frac{3}{4}=\frac{6}{24}\)

\(\Leftrightarrow\left|6x-2\right|=\frac{1}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}6x-2=\frac{1}{3}\\6x-2=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}6x=\frac{7}{3}\\6x=\frac{5}{3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{18}\\x=\frac{5}{18}\end{cases}}\)

\(\left|6x-2\right|\times\frac{3}{4}=\frac{6}{24}\)

<=> \(\left|6x-2\right|=\frac{6}{24}\div\frac{3}{4}\)

<=> \(\left|6x-2\right|=\frac{1}{3}\)

<=> \(\orbr{\begin{cases}6x-2=\frac{1}{3}\\6x-2=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{18}\\x=\frac{5}{18}\end{cases}}\)

23+50.23+50

= 23.(50+50)

= 2300

#Chúchọctốt

\(23+50.23+50\)

\(=\left(23+23\right).\left(50+50\right)\)

\(=56.100\)

\(=5600\)

(-9/4) ^ 3 = -729/64

Gọi các p/s cần tìm là x, theo đề ta có :

\(\frac{1}{3}< x< \frac{4}{7}\)

\(\Rightarrow\frac{7}{21}< x< \frac{12}{21}\)

\(\Rightarrow x\in\left\{\frac{8}{21};\frac{9}{21};\frac{10}{21};\frac{11}{21}\right\}\)

Vậy các p/s cần tìm là \(\frac{8}{21};\frac{9}{21};\frac{10}{21};\frac{11}{21}\)

                                                             Bài giải

             Gọi phân số cần tìm là a 

Ta có : \(\frac{1}{3}< a< \frac{4}{7}\text{ }\Rightarrow\text{ }\frac{7}{21}< a< \frac{12}{21}\)

\(\Rightarrow\text{ }a\in\left\{\frac{8}{21}\text{ ; }\frac{9}{21}\text{ }\frac{10}{21}\text{ ; }\frac{11}{21}\right\}\)

Vậy 4 phân số cần tìm là \(\frac{8}{21}\text{ ; }\frac{9}{21}\text{ }\frac{10}{21}\text{ ; }\frac{11}{21}\)

Tìm GTNN của N?

Ta có: \(N=\left|x-\frac{1}{4}\right|-\frac{4}{5}\)

Mà \(\left|x-\frac{1}{4}\right|\ge0\Rightarrow N=\left|x-\frac{1}{4}\right|-\frac{4}{5}\ge-\frac{4}{5}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(N=-\frac{4}{5}\Leftrightarrow\left|x-\frac{1}{4}\right|=0\)

\(\Leftrightarrow x-\frac{1}{4}=0\Rightarrow x=\frac{1}{4}\)

Vậy GTNN của N là -4/5 khi x = 1/4

Tìm giá trị nhỏ nhất ?

Vì |x - 1,4| \(\ge\)0 \(\forall\)x

=> |x - 1,4| - 4/5 \(\ge\)-4/5 \(\forall\)x

Dấu " = " xảy ra khi và chỉ khi |x - 1,4| = 0 => x = 1,4

Vậy N_min = -4/5 khi x = 1,4

Ta có: 4³⁰=4¹⁵⁺¹⁵=4¹⁵.4¹⁵

           3.24¹⁰=3.(2³.3)¹⁰=3.2³⁰.3¹⁰=2³⁰.3¹¹=(2²)¹⁵.3¹¹=4¹⁵.3¹¹

Mà 4¹⁵=4¹⁵ ; 4¹⁵>3¹¹

=> 4¹⁵.4¹⁵>4¹⁵.3¹¹

=> 4³⁰>3.24¹⁰

=> 2³⁰ + 3³⁰ + 4³⁰>3.24¹⁰

We have: \(1^2+2^2+3^2+...+100^2\)

\(=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+100.\left(101-1\right)\)

\(=\left(1.2+2.3+...+100.101\right)-\left(1+2+...+100\right)\)

\(=\frac{1.2.3+2.3.3+...+100.101.3}{3}-\frac{\left(100+1\right).\left[\left(100-1\right)\div1+1\right]}{2}\)

\(=\frac{1.2.3+2.3.\left(4-1\right)+...+100.101.\left(102-99\right)}{3}-\frac{100.101}{2}\)

\(=\frac{1.2.3-1.2.3+2.3.4-...-99.100.101+100.101.102}{3}-5050\)

\(=\frac{100.101.102}{3}-5050\)

\(=343400-5050\)

\(=338350\)

Từ \(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\frac{d}{a+b+c}\)

\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{d+a+b}+1=\frac{d}{a+b+c}+1\)

Hay \(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{c+d+a}=\frac{a+b+c+d}{d+a+b}=\frac{a+b+c+d}{a+b+c}\)

Nếu a+b+c+d=0 thì \(a+b=-\left(c+d\right)\Rightarrow\frac{a+b}{c+d}=-1;\frac{c+d}{a+b}=-1\)

\(b+c=-\left(d+a\right)\Rightarrow\frac{b+c}{d+a}=-1;\frac{d+a}{b+c}=-1\Rightarrow H=-1\)

Nếu a+b+c+d \(\ne0\)thì: \(b+c+d=c+d+a=d+a+b=a+b+c\Rightarrow a=b=c=d\Rightarrow H=1\)

Vậy H=-1 nếu a+b+c+d=0; H=1 nếu a+b+c+d khác 0

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

=> \(\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{d+a+b}+1=\frac{d}{a+b+c}+1\)

=> \(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)

Nếu a + b + c + d = 0

=> a + b = -(c + d)

=> b + c = -(d + a)

=> c + d = -(b + a)

=> d + a = -(b + c)

Khi đó H = \(\frac{-\left(c+d\right)}{c+d}.\frac{-\left(d+a\right)}{d+a}.\frac{-\left(b+a\right)}{c+d}.\frac{-\left(d+a\right)}{b+c}=\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right)=1\)

Khi a + b + c + d \(\ne\)0

=> \(\frac{1}{b+c+d}=\frac{1}{a+c+d}=\frac{1}{a+b+d}=\frac{1}{a+b+c}\)

=> b + c + d = a + c + d = a + b + d = a + b + c

=> a = b = c = d

Khi đó H = \(\frac{2b}{2d}.\frac{2c}{2a}.\frac{2d}{2b}.\frac{2a}{2c}=1\)