\(x+\dfrac{1}{2}=\dfrac{33}{4}\\ \Rightarrow x=\dfrac{33}{4}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{31}{4}\\ \dfrac{5}{6}-x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{2}\\ x+\dfrac{4}{5}=\dfrac{-2}{3}\\ \Rightarrow x=\dfrac{-2}{3}-\dfrac{4}{5}\\ \Rightarrow x=\dfrac{-22}{15}\)

<=> 2(x^2-25) - 2x^2+3x-4x+6 + x^3-8x = x^3+1

=>2x^2-50 - 2x^2 -9x+6+x^3-x^3-1 = 0

<=>-9x - 45 =0

<=>-9x=45

<=>x=-5

Còn phần b và c bạn cứ khai triển ra,mình phải đi học nên không có thời gian giải cho bạn

`Answer:`

\(3\left(\frac{5}{3}x-7\right)-2\left(1.5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow3\left(\frac{5x}{3}-7\right)-2\left(5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+x^2+4x=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+4x=80\)

\(\Leftrightarrow-6x-53=80\)

\(\Leftrightarrow-6x=133\)

\(\Leftrightarrow x=-\frac{133}{6}\)

\(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)\left(4\frac{x^2}{3}+1\right)=\frac{22}{45}x^2\)

\(\Leftrightarrow36x^2\left(\frac{x}{3}-\frac{1}{2}\right)-45\left(\frac{x}{5}-\frac{2}{3}\right)\left(\frac{4x^2}{3}+1\right)=22x^2\)

\(\Leftrightarrow12x^3-18x^2-12x^3-9x+40x^2+30=22x^2\)

\(\Leftrightarrow22x^2-9x+30=22x^2\)

\(\Leftrightarrow-9x+30=0\)

\(\Leftrightarrow-9x=-30\)

\(\Leftrightarrow x=\frac{10}{3}\)

1).( 27,56 x 35 ) + ( 27,56 x 67 ) - ( 27,56 x 2)

= (964 + 1846,52) - 55,12

=2810,52 - 55,12

= 2755,4

2).( 4x 35 ) x ( 25 x 5 ) x 2

= ( 140 x 125 ) x2

= 17500 x 2

=35000

4). 3/10

5). 1188

6).  61/6

1)2756   2)35000   3)0   4)3/10   5)10/1/6

1,2755,4

2,35000

3,0

4,\(\frac{3}{10}\)

5,1188

6,\(\frac{61}{6}\)

1.

$x(x+2)(x+4)(x+6)+8$

$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$

$=a(a+8)+8$ (đặt $x^2+6x=a$)

$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$

Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$

2.

$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$

$=5-(x^2+5x-6)(x^2+5x+6)$

$=5-[(x^2+5x)^2-6^2]$

$=41-(x^2+5x)^2\leq 41$

Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$

3.

Đặt $x+3=a; 7-x=b$ thì $a+b=10$ 

$C=a^4+b^4$

Áp dụng BĐT Bunhiacopxky:

$(a^4+b^4)(1+1)\geq (a^2+b^2)^2$

$\Rightarrow C\geq \frac{(a^2+b^2)^2}{2}$
$(a^2+b^2)(1+1)\geq (a+b)^2=100$

$\Rightarrow a^2+b^2\geq 50$

$\Rightarrow C\geq \frac{50^2}{2}=1250$

Vậy $C_{\min}=1250$

Giá trị này đạt tại $a=b=5\Leftrightarrow x=2$

\(6-2\left(x-1\right)=4\)

\(\Rightarrow2\left(x-1\right)=6-4\)

\(\Rightarrow2\left(x-1\right)=2\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=1+1=2\)

________________

\(2\cdot\left(x-2\right)+1=7\)

\(\Rightarrow2\cdot\left(x-2\right)=7-1\)

\(\Rightarrow2\cdot\left(x-2\right)=6\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=3+2=5\)

_______________

\(\left(2\cdot x-3\right)+4=9\)

\(\Rightarrow2\cdot x-3=5\)

\(\Rightarrow2\cdot x=3+5\)

\(\Rightarrow2\cdot x=8\)

\(\Rightarrow x=\dfrac{8}{2}=4\)

________________

\(\left(3\cdot x-2\right)-1=3\)

\(\Rightarrow3\cdot x-2=3+1\)

\(\Rightarrow3\cdot x-2=4\)

\(\Rightarrow3\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{3}=2\)

a: =>2(x-1)=2

=>x-1=1

=>x=2

b: =>2(x-2)=6

=>x-2=3

=>x=5

c; =>2x-3=5

=>2x=8

=>x=4

d: =>3x-2=4

=>3x=6

=>x=2

e: =>2(6-x)=4

=>6-x=2

=>x=4

f: =>x-2=5

=>x=7

g: =>10-2x=4

=>2x=6

=>x=3

h: =>2x+4=3

=>2x=-1

=>x=-1/2

j: =>x+2=12

=>x=10

l: =>2x+3=3

=>2x=0

=>x=0

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

\(2\left(6-x\right)+1=5\\ 2\left(6-x\right)=5-1=4\\ 6-x=\dfrac{4}{2}=2\\ x=6-2=4\)

\(\left(4+2x\right)-1=8\\ 3\left(4+2x\right)=8+1=9\\ 4+2x=\dfrac{9}{3}=3\\ 2x=3-4=-1\\ x=-\dfrac{1}{2}\)

\(2\left(x-2\right)+1=7\\ 2\left(x-2\right)=7-1=6\\ x-2=\dfrac{6}{2}=3\\ x=3+2=5\)

\(20+\left(x-2\right)=25\\ x-2=25-20=5\\ x=5+2=7\)

\(\left(x+2\right)-5=7\\ x+2=7+5=12\\ x=12-2=10\)

\(\left(2x-3\right)+4=9\\ 2x-3=9-4=5\\ 2x=5+3=8\\ x=\dfrac{8}{2}=4\)

\(\left(10-\dfrac{2}{x}\right)-1=3\\ 10-\dfrac{2}{x}=3+1=4\\ \dfrac{2}{x}=10-4=6\\ x=\dfrac{2}{6}=\dfrac{1}{3}\)

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

\(\left(3x-2\right)-1=3\\ 3x-2=3+1=4\\ 3x=4+2=6\\ x=\dfrac{6}{3}=2\)

\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)

a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)

ĐKXĐ: x ≠ -1

⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)

⇔ 65 + 52 = -3(x + 1)

⇔ 117 = -3x - 3

⇔ 117 + 3 = -3x

⇔ 120 = -3x 

⇔ x = \(\dfrac{120}{-3}=-40\) (TM)

b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)

⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)

⇔ 4x = -2,75

⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)

c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)

⇔  \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)

⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x² + 96x + 144x + 48

⇔ 936x + 1560x - 624x - 96x - 144x - 288x² = 48 - 312 - 520 + 312

⇔ 1632x - 288x² = -472

⇔ -288x² + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)

⇔ x = 5,942459684 \(\approx\) 6

c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow2x+1=13\)

hay x=6