Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Rightarrow\left(x-2\right)\left(x-2\right)-\left[x^2-3^2\right]=6\)
\(\Rightarrow\left(x^2-4x+4\right)-x^2+9=6\)
\(\Rightarrow-4x+\left(4+9\right)=6\)
\(\Rightarrow-4x+13=6\)
\(\Rightarrow-4x=6-13\)
\(\Rightarrow-4x=-7\)
\(\Rightarrow x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)
~ Ủng hộ nhé
( x - 2 )2 - ( x - 3 ) ( x + 3 ) = 6
x2 - 4x + 4 - x2 + 9 = 6
( x2 - x2 ) - 4x + ( 4 + 9 ) = 6
-4x + 13 = 6
-4x = 6 - 13
-4x = -7
x = 7/4
\(\text{a , (x-3).(x^2+3x+9)+x(x+2).(2-x)=1 }\)
=(x3-33)+x(4-x2)=1
=x3-27+4x-x3=1
4x-27=1
4x=28
x=7
\(\text{b, (x+1)^3-(x-1)^3-6.(x-1)^2=-10}\)
=-0,5
\(\left(x+2\right)\left(x-3\right)-\left(x-3\right)^2=15\)
\(\Leftrightarrow\left(x+2-x+3\right)\left(x-3\right)=15\)
\(\Leftrightarrow5\left(x-3\right)=15\)
\(\Leftrightarrow x-3=3\)
\(\Leftrightarrow x=6\)
b (x+1)^3-(x^3+3x^2+2x-3)=6
\(\Leftrightarrow x^3+3x^2+3x+1-x^3-3x^2-2x+3=6\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
( x - 2 )2 - ( x - 3 )( x + 3 ) = 6
<=>(x-2)2-(x-3)2=6
<=>(x-2-x+3)(x-2+x-3)=6
<=>2x-5=6
<=>2x=11
<=>x=\(\frac{11}{2}\)
( x - 2 )2 - ( x - 3 )( x + 3 ) = 6
<=>x2-4x+4-x2+9=6
<=>-4x=-7
<=>x=\(\frac{7}{4}\)
\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(x^2-4x+4-x^2+9=6\)
\(-4x+13=6\)
\(-4x=6-13\)
\(-4x=-7\)
\(x=\frac{-7}{-4}\)
\(x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)
Bài giải
\(x^2-4x+4-x^2+9=6\)
\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(-4x+13=6\)
\(-4x=6-13\)
\(-4x=-7\)
\(x=\frac{-7}{-4}\)
\(x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)