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10 tháng 7 2020

Trả lời 

\(x.\left(x+2\right).\left(x+4\right).\left(x+6\right)=9\)

\(\Leftrightarrow\left[x.\left(x+6\right)\right].\left[\left(x+2\right).\left(x+4\right)\right]=9\)

\(\Leftrightarrow\left(x^2+6x\right).\left(x^2+6x+8\right)=9\)

Đặt \(x^2+6x=t\) ta có 

\(t.\left(t+8\right)=9\)

\(\Leftrightarrow t^2+8t-9=0\)

\(\Leftrightarrow t^2-t+9t-9=0\)

\(\Leftrightarrow t.\left(t-1\right)+9.\left(t-1\right)=0\)

\(\Leftrightarrow\left(t-1\right).\left(t+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-1=0\\t+9=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}t=1\\t=-9\end{cases}}\)

TH1 \(t=1\)

\(\Rightarrow x^2+6x=1\)

\(\Leftrightarrow x^2+6x-1=0\)

\(\Leftrightarrow x^2+6x+9-10=0\)

\(\Leftrightarrow\left(x+3\right)^2=10=\left(\pm\sqrt{10}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=\sqrt{10}\\x+3=-\sqrt{10}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3+\sqrt{10}\\x=-3-\sqrt{10}\end{cases}}\)

TH2: \(t=-9\)

\(\Rightarrow x^2+6x=-9\)

\(\Leftrightarrow x^2+6x+9=0\)

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

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy \(x\in\left\{-3+\sqrt{10};-3-\sqrt{10};-3\right\}\)

\(x\left(x+2\right)\left(x+4\right)\left(x+6\right)=9\)

\(\Leftrightarrow x^4+12x^3+44x^2+48x=9\)

\(\Leftrightarrow x^4+12x^3+44x^2+48x-9=0\)

\(\Leftrightarrow\left(x^3+9x^2+17x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2+6x-1\right)\left(x+3\right)^2=0\)

TH1 : Ta có : \(6^2-4.\left(-1\right)=36+4=40>0\)Suy ra : \(x_1=\frac{-6-\sqrt{40}}{2};x_2=\frac{-6+\sqrt{40}}{2}\)

TH2 : \(\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

10 tháng 12 2015

a)\(x^2-4^2+6x-x^2=0\)

\(16+6x=0\)

\(x=\frac{8}{3}\)

b)x=3

12 tháng 6 2023

\(\left(x+9\right)+\left(x-2\right)+\left(x+7\right)+\left(x-4\right)+\left(x+5\right)+\left(x-6\right)+\left(x+3\right)+\left(x-8\right)+\left(x+1\right)=95\)

\(x+9+x-2+x+7+x-4+x+5+x-6+x+3+x-8+x+1=95\)

\(\left(x\times9\right)+\left(9-2+7-4+5-6+3-8+1\right)=95\)

\(x\times9+5=95\)

\(x\times9=95-5\)

\(x\times9=90\)

\(x=90:9\)

\(x=10\)

12 tháng 6 2023

(x + 9) + (x - 2) + (x + 7) + (x - 4) + (x + 5) + (x - 6) + (x + 3) + (x - 8) + (x + 1) = 95

x + 9 + x - 2 + x + 7 + x - 4 + x + 5 + x - 6 + x + 3 + x - 8 + x + 1

(x + x + x + x + x + x + x + x + x) + (9 - 8 + 7 - 6 + 5 - 4 + 3 - 2 + 1) = 95

9 × x + 5 = 95

9 × x = 95 - 5

9 × x = 90

x = 90 : 9

x = 10

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

18 tháng 4 2022

a)4/9 : x/6 = 2/3

           x/6 = 4/9 : 2/3

          x/6  =   2/3 = 4/6

vậy x = 4

b)x/5 x 3/7 = 9/35

   x/5          = 9/35 : 3/7

   x/5          = 3/5

vậy x = 3

16 tháng 5 2023

rảnh à

 

6 tháng 11 2023

100

26 tháng 1 2023

\(\left(x+9\right)+\left(x-8\right)+\left(x+7\right)+\left(x-6\right)+\left(x+5\right)+\left(x-4\right)+\left(x+3\right)+\left(x-2\right)+\left(x+1\right)=95.9\\ =>x+9+x-8+x+7+x-6+x+5+x-4+x+3+x-2+x+1=855\\ =>\left(x+x+x+x+x+x+x+x+x\right)+\left(9-8+7-6+5-4+3-2+1\right)=855\\ =>9x+5=855\\ =>9x=855-5\\ =>9x=850\\ =>x=\dfrac{850}{9}\)

26 tháng 1 2023

\(\left(x+9\right)+\left(x-8\right)+...+\left(x-2\right)+\left(x+1\right)\)

\(=x+9+x-8+...+x-2+x+1\)

\(=\left(x+9+x-8\right)+...+\left(x+5\right)+...+\left(x-2+x+1\right)\)

(Ta gộp 4 số vào 1 tổng, riêng (x+5) là ta giữ nguyên)

\(=\left(2x-1\right)+...+\left(x+5\right)+...+\left(2x-1\right)\)

\(=4\left(2x-1\right)+\left(x+5\right)\)

\(=8x-4+x-5\)

\(=9x-9\)        (1)

Từ bài toán trên, ta có:

\(\left(x+9\right)+\left(x-8\right)+...+\left(x-2\right)+\left(x+1\right)=95,9\)

Từ (1)

\(\Leftrightarrow9x-9=95,9\)

\(9x=95,9-9\)

\(x=86,9:9\)

\(x=9,6\left(5\right)\)

 

a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)

hay x=0

Vậy: x=0

b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)

\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)

\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)

hay \(x=\dfrac{-1}{9}\)

Vậy: \(x=\dfrac{-1}{9}\)