`24,45+41,2+75,55+8,8=(24,45+75,55)+(41,2+8,8)=100+50=150`

`4,5xx9+4,5=4,5xx(9+1)=4,5xx10=45`

tính bằng cách thuận tiện nhất :
24,45 + 41,2 + 75,55 + 8,8

= ( 24,45 + 75,55 ) + ( 41,2 + 8,8 )

= 100 + 50

= 150.

4,5 x 9 + 4,5

= ( 4,5 x 9 ) + 4,5

= 40,5 + 4,5

= 45.

24,45 + 41,2 + 75,55 + 8,8

= (41,2 + 8,8) + (24,45 + 75,55)

= 50 + 100

= 150

4,5 x 9 + 4,5

= 9 x (4,5 + 4,5 + 1)

= 9 x 10

= 90

\(B=2\left|4,5x-9\right|-18\)

Vì \(\left|4,5x-9\right|\ge0\forall x\)

=> \(2\left|4,5x-9\right|-18\ge-18\)

Dấu " = " xảy ra khi và chỉ khi |4,5x - 9| = 0 => 4,5x - 9 = 0 => 4,5x = 9 => x = 2

Vậy \(B_{min}=-18\)khi x = 2

\(C=\left(2x+1\right)^2-1990\)

Vì \(\left(2x+1\right)^2\ge0\forall x\)

=> \(\left(2x+1\right)^2-1990\ge-1990\forall x\)

Dấu " = " xảy ra khi và chỉ khi (2x + 1)² = 0 => 2x + 1 = 0 => x = -1/2

Vậy \(C_{min}=-1990\)khi x = -1/2

\(D=\left(x+1\right)^2+\left|y+5\right|-\frac{3}{2}\)

Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left|y-5\right|\ge0\forall y\end{cases}}\)

=> \(\left(x+1\right)^2+\left|y+5\right|\ge0\forall x\)

=> \(\left(x+1\right)^2+\left|y+5\right|-\frac{3}{2}\ge-\frac{3}{2}\forall x\)

Dấu " = " xảy ra khi và chỉ khi \(\hept{\begin{cases}\left(x+1\right)^2=0\\\left|y+5\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=-5\end{cases}}\)

Vậy \(D_{min}=-\frac{3}{2}\)khi \(\hept{\begin{cases}x=-1\\y=-5\end{cases}}\)

a: =>2x-1=-2

=>2x=-1

hay x=-1/2

b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)

c: x/8=9/4

nên x/8=18/8

hay x=18

d: \(\Leftrightarrow\left(x-3\right)^2=36\)

=>x-3=6 hoặc x-3=-6

=>x=9 hoặc x=-3

e: =>-1,7x=6,12

hay x=-3,6

h: =>x-3,4=27,6

hay x=31

a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)

\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)

\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)

\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)

\(\dfrac{1}{3}=-2x+1\div6\)

\(x=-\dfrac{1}{2}\)

b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)

\(TH1:3x+2=0\)

\(3x=0-2\)

\(3x=-2\)

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

\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)

\(\left(\dfrac{-2}{5}x-7\right)=0\)

\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)

\(\left(\dfrac{-2x-35}{5}\right)=0\)

\(-2x-35=0\)

\(-2x=0+35\)

\(x=-\dfrac{35}{2}\)

c) \(\dfrac{x}{8}=\dfrac{9}{4}\)

\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)

\(x=18\)

d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)

\(x-3=18+2\)

\(x=20-3\)

\(x=17\)

e) \(4,5x-6,2x=6,12\)

\(\dfrac{9x}{2}-6,2.x=6,12\)

\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)

\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)

\(\dfrac{45x-62x}{10}=6.12\)

\(=-17x\div10=6.12\)

\(-17x=10.6.12\)

\(x=-3,6\)

h) \(11,4-\left(x-3,4\right)=-16,2\)

\(x-3,4=-16,2+11,4\)

\(x-3,4=-4,8\)

\(x=-1,4\)

a) x⁴ + 3x² + 4

= [ ( x² )² + 4x² + 2² ] - x²

= ( x² + 2 )² - x²

= ( x² + 2 - x² ).( x² + 2 + x² )

= 2( 2x² + 2 )

b) x⁴ + 5x² + 9

= [ ( x² )² + 6x² + 3² ] - x²

= ( x² + 3 )² - x²

= ( x² + 3 - x² )( x² + 3 + x² )

= 3( 2x² + 3 )

a) (x + 1)² = 4/3. 75/9

=> (x + 1)² = 100/9

=> (x + 1)² = (10/3)²

=> \(\orbr{\begin{cases}x+1=\frac{10}{3}\\x+1=-\frac{10}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{13}{3}\end{cases}}\)

b) (4,5x - 2x) . (-11/7) = 11/14

=> 2,5x = 11/14 : (-11/7)

=> 2,5x = -1/2

=> x = -1/2 : 2,5

=> x = -0,2

#)Giải :

Tìm số tự nhiên X :

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}:2=\frac{2001}{2003}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(\Rightarrow x=2003-1\)

\(\Rightarrow x=2002\)

                #~Will~be~Pens~#

\(\left(2x-5\right)\left(x-3\right)+\left(2x-5\right)^2=0\)

\(\Rightarrow\left(2x-5\right)\left(x-3+2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(3x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\3x-8=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{8}{3}\end{cases}}\)

\(\frac{3x-5}{4}+\frac{2x-3}{6}=\frac{x}{3}-1\)

\(\Leftrightarrow\frac{18x-30+8x-12}{24}=\frac{x-3}{3}\)

\(\Leftrightarrow\frac{26x-42}{24}=\frac{x-3}{3}\)

\(\Leftrightarrow78x-126=24x-72\)

Chuyển vế các kiểu

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

\(\Leftrightarrow\left(2x-6\right)\left(3x-2\right)+3x\left(2x-5\right)=2\left(2x-5\right)\left(3x-2\right)\)

\(\Leftrightarrow6x^2-4x-18x+12+6x^2-15x=2\left(6x^2-4x-15x+10\right)\)

\(\Leftrightarrow12x^2-27x+12=12x^2-38x+20\)

Đến đây EZ rồi

   4,2 \(\times\)(-5,1) - 5,1 \(\times\) (-5,8) + (-0,16)

=  5,1\(\times\)(- 4,2 + 5,8) - 0,16

= 5,1 \(\times\) 1,6 - 0,16

= 8,16 - 0,16 

= 8 

= 8

 \(\text{4,2 x (-5,1) - 5,1 x (-5,8) + (-0,16)}\) \(\text{=}\)

\(\text{ -21,42 - (-29,58) + (-0,16)}\)\(\text{=}\)

\(\text{ 8,16 + (-0,16) }\) \(\text{=}\)

 \(\text{8}\)

\(4,5x+\dfrac{1}{2}-x=1\dfrac{1}{3}\)

\(3,5x=1\dfrac{1}{3}-\dfrac{1}{2}\)

\(3,5x=\dfrac{5}{6}\)

\(x=\dfrac{5}{6}:3,5=\dfrac{5}{21}\)