Cái này chỉ có GTNN

`(3-2x)^2>=0`

`=>12(3-2x)^2>=0`

Dấu "=" xảy ra khi `3-2x=0<=>2x=3<=>x=3/2.`

xin đếy giải giúp mik ik!!!

a,21x6+18x6+6x61

=6x(21+18+61)

=6x100

=600

b,1078x25-25x35-43x25

=25x(1078-35-43)

=25x1000

=25000

c,74x18+740x6+22x74

=74x18+74x10x6+22x74

=74x18+74x60+22x74

=74x(18+60+22)

=74x100

=7400

a) x : 1/4 = 7/5
    x         = 7/5 x 1/4
    x         = 7/20
b) 9/2 - x = 3/5
            x = 9/2 - 3/5
            x = 39/10
c) 12/5 : x = 7/2
              x = 12/5 : 7/2
              x = 24/35

a) \(x\div\dfrac{1}{4}=\dfrac{7}{5}\)

    \(x=\dfrac{7}{5}\times\dfrac{1}{4}\)

    \(x=\dfrac{7}{20}\)

b) \(\dfrac{9}{2}-x=\dfrac{3}{5}\)

            \(x=\dfrac{9}{2}-\dfrac{3}{5}\)

            \(x=\dfrac{39}{10}\)

c) \(\dfrac{12}{5}\div x=\dfrac{7}{2}\)

             \(x=\dfrac{12}{5}\div\dfrac{7}{2}\)

             \(x=\dfrac{12}{5}\times\dfrac{2}{7}\)

             \(x=\dfrac{24}{35}\)

a)\(x=\dfrac{1}{4}\)x\(\dfrac{7}{5}=\dfrac{7}{20}\)

b)\(x=\dfrac{9}{2}-\dfrac{3}{5}=\dfrac{39}{10}\)

c)\(x=\dfrac{12}{5}:\dfrac{7}{2}=\dfrac{24}{35}\)

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

\(=\left(2ab-a^2-b^2+c^2\right)\left(2ab+a^2+b^2-c^2\right)\)

\(=-\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]\)

\(=-\left(a-b-c\right)\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)\)

\(A\left(x\right)\) chia hết cho \(B\left(x\right)\) khi \(A\left(2\right)=0\)

\(\Rightarrow3.2^2+5.2+m=0\)

\(\Rightarrow m=-22\)

\(5n+14⋮n+2\)

\(\Leftrightarrow n+2\in\left\{2;4\right\}\)

hay \(n\in\left\{0;2\right\}\)

Để 5n+14 chia hết n+2
<=> 2(5n+14) chia hết n+2
<=> 10n + 28 chia hết n+2
<=> 10n+20+8 chia hết n+2
<=> 8 chia hết n+2
<=> n+2 thuộc Ư(8) = {1; 2; 4}
<=> n thuộc {-1; 0; 2}
mà n thuộc N
=> n thuộc {2; 0}

\(M=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{4^2}< \dfrac{1}{3\cdot4};...;\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

\(\Rightarrow M< \dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-1-\dfrac{1}{2}-...-\dfrac{1}{50}\\ =\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\left(50.số\right)=\dfrac{50}{50}=1\)

Vậy \(M< 1\)

Mình chỉ so sánh với 1 được thôi à :((

b.\(\left(\dfrac{1}{2}-\dfrac{3}{2x-5^2-5}\right)=11\)

\(\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{3}{2x-30}\right)=11\);\(x\ne15\)

\(\Leftrightarrow-\dfrac{3}{2x-30}=\dfrac{21}{2}\)

\(\Leftrightarrow-\dfrac{3}{2\left(x-15\right)}=\dfrac{21}{2}\)

\(\Leftrightarrow-\dfrac{3}{2\left(x-15\right)}=\dfrac{21\left(x-15\right)}{2\left(x-15\right)}\)

\(\Leftrightarrow-3=21\left(x-15\right)\)

\(\Leftrightarrow-3=21x-315\)

\(\Leftrightarrow21x=312\)

\(\Leftrightarrow x=\dfrac{104}{7}\)

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

\(\Leftrightarrow-\left|4x+\dfrac{1}{2}\right|=-\dfrac{1}{6}\)

\(\Leftrightarrow\left|4x+\dfrac{1}{2}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\dfrac{1}{2}=\dfrac{1}{6}\\4x+\dfrac{1}{2}=-\dfrac{1}{6}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{12}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

a.\(\left(\dfrac{1}{2}-\dfrac{3}{2x-5^2-5}\right)=11\)

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

Đề là vậy đk bạn?