\(A=\left(37,1-4,5\right)-\left(-4,5+37,1\right)\)

   \(=37,1-4,5+4,5-37,1\)

   \(=37,1-37,1-4,5+4,5\)

   \(=0.\)

\(B=-\left(315.4+275\right)+4.315-\left(10-275\right)\)

   \(=-315.4-275+4.315-10+275\)

   \(=-315.5+4.315-275+275-10\)

   \(=0+0-10=-10.\)

\(C=-\left(\frac{3}{7}+\frac{3}{8}\right)-\left(-\frac{3}{8}+\frac{4}{7}\right)\)

    \(=-\frac{3}{7}-\frac{3}{8}+\frac{3}{8}-\frac{4}{7}\)

    \(=-\frac{3}{7}-\frac{4}{7}-\frac{3}{8}+\frac{3}{8}\)

    \(=-1+0=-1.\)

A=37,1 - 4,5 + 4,5 - 37,1

A=(37,1 - 37,10) + ( 4,5 - 4,5 )

 A = 0 + 0 = 0

B= -315.4 - 275  + 4.315 - 10 + 275

B=(-315.4 + 4.315) + ( 275-275) - 10

B= 0 + 0 - 10 = -10

C= -3/7 - 3/8 + 3/8 - 4/7

C = ( -3/7-4/7) + ( 3/8 - 3/8)

C=-7/7 + 0 = -7/7 = -1

a/ \(\Leftrightarrow2x^3+9x^2-27=0\)

\(\Leftrightarrow2x^3+12x^2+18x-3x^2-18x-27=0\)

\(\Leftrightarrow2x\left(x^2+6x+9\right)-3\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+3\right)^2=0\)

\(\Leftrightarrow...\)

b/ \(\Leftrightarrow x^3-3x^2+3x-1+x^3+x^3+3x^2+3x+1=x^3+6x^2+12x+8\)

\(\Leftrightarrow x^3-3x^2-3x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2+x+1\right)=0\)

c/ \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt \(x^2+x=t\)

\(t\left(t-2\right)-24=0\Leftrightarrow t^2-2t-24=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+x=6\\x^2+x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+4=0\end{matrix}\right.\)

d/ \(\Leftrightarrow\left(x-7\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)-72=0\)

\(\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\)

Đặt \(x^2-9x+14=0\)

\(t\left(t+6\right)-72=0\Leftrightarrow t^2+6t-72=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-9x+14=6\\x^2-9x+14=-12\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-9x+8=0\\x^2-9x+26=0\end{matrix}\right.\)

Ta có \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Để tổng trên chia hết cho 81 thì \(\left(a-b\right)\left(b-c\right)\left(c-a\right)⋮27\)

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

Bài toán trở thành: Cho \(x+y+z=\left(x-y\right)\left(y-z\right)\left(z-x\right)\). CMR: \(x+y+z⋮27\) - Hoc24

wow

wow, chắc xu học lớp 9

\(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(a+b\right)\left(b+c\right)}{64}}=\dfrac{3a}{4}\)

Tương tự:

\(\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{c+a}{8}\ge\dfrac{3b}{4}\)

\(\dfrac{c^3}{\left(c+a\right)\left(a+b\right)}+\dfrac{c+a}{8}+\dfrac{a+b}{8}\ge\dfrac{3c}{4}\)

Cộng vế:

\(VT+\dfrac{4\left(a+b+c\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow VT\ge\dfrac{a+b+c}{4}\)

Dấu "=" xảy ra khi \(a=b=c\)

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

\(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{8}\\x=\frac{11}{8}\end{matrix}\right.\)

Vậy....

b)\(7,5-3\left|5-2x\right|=-4,5\)

\(\Leftrightarrow\left|5-2x\right|=4\)

\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)

VẬy...

c)\(\left|3x-4\right|+\left|5-2x\right|=0\)

Có: \(\left|3x-4\right|\ge0với\forall x\\ \left|5-2x\right|\ge0với\forall x\)

\(\Rightarrow\left[{}\begin{matrix}3x-4=0\\5-2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=\frac{5}{2}\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)

Bài 3 :

\(a,\left(\frac{2}{5}\right)^2+5\frac{1}{2}:\left(4,5-2\right)+\frac{2^3}{\left(-4\right)}\)

\(=\frac{4}{25}+\frac{11}{2}:\frac{5}{2}+\frac{8}{\left(-4\right)}\)

\(=\frac{4}{25}+\frac{11}{5}-2\)

\(=\frac{59}{25}-2\)

\(=\frac{9}{25}\)

b, Giống câu a

Học tốt

a) \(\overrightarrow{AB}\left(2;-2\right)\); \(\overrightarrow{CA}=\left(4;-4\right)\).
Vì \(\dfrac{2}{4}=\dfrac{-2}{-4}\) nên \(\overrightarrow{AB};\overrightarrow{CA}\) cùng phương . Suy ra ba điểm A, B, C thẳng hàng.
\(\overrightarrow{AB}\left(2;1\right)\); \(\overrightarrow{AC}\left(m+3;2m\right)\).
3 điểm A, B, C thẳng hàng nên hai véc tơ \(\overrightarrow{AB},\overrightarrow{AC}\) cùng phương.
Suy ra: \(\dfrac{m+3}{2}=\dfrac{2m}{1}\Leftrightarrow m+3=4m\)\(\Leftrightarrow m=1\).

Dấu >= hay <= vậy bạn? Bạn xem lại đề.

\(\dfrac{a^3}{\left(a+2b\right)\left(b+2c\right)}+\dfrac{a+2b}{27}+\dfrac{b+2c}{27}\ge3\sqrt[3]{\dfrac{a^3\left(a+2b\right)\left(b+2c\right)}{27^2.\left(a+2b\right)\left(b+2c\right)}}=\dfrac{a}{3}\)

Tương tự:

\(\dfrac{b^3}{\left(b+2c\right)\left(c+2a\right)}+\dfrac{b+2c}{27}+\dfrac{c+2a}{27}\ge\dfrac{b}{3}\)

\(\dfrac{c^3}{\left(c+2a\right)\left(a+2b\right)}+\dfrac{c+2a}{27}+\dfrac{a+2b}{27}\ge\dfrac{c}{3}\)

Cộng vế:

\(VT+\dfrac{2\left(a+b+c\right)}{9}\ge\dfrac{a+b+c}{3}\)

\(\Rightarrow VT\ge\dfrac{a+b+c}{9}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)