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\(\Leftrightarrow x^2-36-x^2+12x-9=9\)
\(\Leftrightarrow12x=54\)
hay x=9/2
Ta có
( x – 6 ) ( x + 6 ) – ( x + 3 ) 2 = 9 ⇔ x 2 – 36 – ( x 2 + 6 x + 9 ) = 9 ⇔ x 2 – 36 – x 2 – 6 x – 9 – 9 = 0
ó - 6x – 54 = 0 ó 6x = -54 ó x = -9
Vậy x = -9
Đáp án cần chọn là: A
a) \(\dfrac{x}{12}=\dfrac{6}{9}\)
\(\Rightarrow x=\dfrac{12\cdot6}{9}=8\)
b) \(\dfrac{9}{15}=x+\dfrac{11}{35}\)
\(\Rightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\)
\(\Rightarrow x=\dfrac{2}{7}\)
c) \(x+\dfrac{6}{5}=\dfrac{x}{2}\)
\(\Rightarrow2\left(x+\dfrac{6}{5}\right)=x\)
\(\Rightarrow2x+\dfrac{12}{5}=x\)
\(\Rightarrow2x-x=-\dfrac{12}{5}\)
\(\Rightarrow x=-\dfrac{12}{5}\)
\(a,\dfrac{x}{12}=\dfrac{6}{9}\\ \Leftrightarrow x=12\cdot\dfrac{6}{9}\\ \Leftrightarrow x=8\\ b,\dfrac{9}{15}=x+\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{2}{7}\\ c,x+\dfrac{6}{5}=\dfrac{x}{2}\\ \Leftrightarrow\dfrac{x}{2}=-\dfrac{6}{5}\\ \Leftrightarrow x=-\dfrac{12}{5}\)
Bài làm
( 3 + 6 + x ) . 9 = ( 2 - x ) . 2
27 + 54 + 9x = 4 - 2x
9x + 2x = 4 - 27 - 54
11x = -77
x = -7
Vậy x = -7
b) ( 6 - x ) . 3 = ( 9 + x ) . 3
18 - 3x = 27 + 3x
-3x - 3x = 27 - 18
-6x = 9
x = 9/-6
x = -3/2
Vậy x = -3/2
a,\(\text{(3+6+x) . 9 = (2-x).2}\)
\(27+54+9x=4-2x\)
\(9x+2x=4-54-27\)
\(11x=-77\)
\(\Rightarrow x=-7\)
b, \(\text{(6-x) . 3 = (9+x) . 3}\)
\(18-3x=27+3x\)
\(-3x-3x=27-18\)
\(-6x=9\)
\(\Rightarrow x=\frac{9}{-6}\)
học tốt
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 : \(\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
134-2{156-6[54-2.(9+6)]}.x=80
=>2.{156-6.[54-2.15]}.x=134-80
=>2.{156-6.[54-30]}.x=54
=>{156-6.24}.x=54:2
=>{156-144}.x=27
=>12.x=27
=>x=27/12
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
134 - 2 . {156 - 6[54 - 2(9 + 6)]} . x = 86
134 - 2 . { 156 - 6 [ 54 - 2 . 15 ) ] } . x = 86
134 - 2 . { 156 - 6 [ 54 - 30 ] } . x = 86
134 - 2 . { 156 - 6 . 24 } . x = 86
134 - 2 . { 156 - 144 } . x = 86
134 - 2 . 21 . x = 86
2 . 21 . x = 134 - 86
2 . 21 . x = 48
21 . x = 48 : 2
21 . x = 24
x = 24 : 21
x = 1
TL
x=3
HT
Sửa đề :
x : 2 = 9 : 68
=> x : 2 = 32 : ( 3 . 2 )8
=> x : 2 = 32 : 38 : 28
=> x : 2 = \(\frac{1}{3^6.2^8}\)
\(\Rightarrow x=\frac{1}{3^6.2^8}.2\)
\(\Rightarrow x=\frac{1}{3^6.2^7}=\frac{1}{729.128}=\frac{1}{93312}\)