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b)(x+3)\(^2\)-x\(^2\)=9
(x+3-x)(x+3+x)=9
3(2x+3)=9
2x+3=3
2x=0
x=0
c)\(x^2+4x+4+x^2-6x+9-2x^2+2=9\)
-2x+15=9
-2x=-6
x=3
a) x(x-2) - 5x + 10 = 0
x(x - 2) - 5(x - 2)=0
(x - 2)(x - 5)=0
=> x=2 ; x=5
Bài 1 :
\(a)\)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+3\right)\left(x-3\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x\left(x^2-3^2\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x^3+9x=15\)
\(\Leftrightarrow\)\(9x=16\)
\(\Leftrightarrow\)\(x=\frac{16}{9}\)
Vậy \(x=\frac{16}{9}\)
Chúc bạn học tốt ~
a) 4x2 - 9=0
(2x)2 - 32 = 0
=》(2x - 3)(2x+3) =0
=》 2x - 3 = 0 hoặc 2x +3 = 0
=》x = 1,5 hoặc x = - 1,5
b) (x + 1)2 - 16 = 0
=》( x + 1)2 - 42 = 0
=》( x - 3 )( x + 5 ) =0
=》 x - 3 = 0 hoặc x + 5 = 0
=》 x = 3 hoặc x = -5
c) ( x + 1)2 - (2x + 3)2 = 0
=》 ( x + 1 - 2x - 3)(x+1 +2x +3 ) =0
=》 ( -x - 2 )( 3x + 4 ) = 0
=》 -x -2 =0 hoặc 3x + 4 = 0
=》 x = -2 hoặc x = -4/3
d) 4(3x +2)2 - 9( x + 1 )2 =0
=》 [ 2(3x +2) ]2 - [3 (x + 1)] 2 = 0
=> ( 6x +4 )2 - ( 3x + 3)2 = 0
=》 ( 6x +4 -3x -3 )( 6x + 4 + 3x + 3 )=0
=》 (3x +1)(9x + 7 ) =0
=》 3x + 1 =0 hoặc 9x + 7 =0
=》 x = -1/3 hoặc x = -7/9
a)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x^3+2^3\right)-x^3-2x=0\)
\(\Leftrightarrow8-2x=0\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
b)\(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
\(x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(x^3-3x^2+3x-1-x^3-27+3x^2-12-2=0\)
\(3x-42=0\)
\(3x=42\)
\(x=14\)
I don't now
or no I don't
..................
sorry
a) \(x^4-x^3-7x^2+x+6=0\)
\(\Leftrightarrow\)\(x^4-x^3-7x^2+7x-6x+6=0\)
\(\Leftrightarrow\)\(x^3\left(x-1\right)-7x\left(x-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^3-7x-6\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=0\)
đến đây lm tiếp
Bài 1:
a) \(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)
\(114x^2+216x+81=114x^2-480x+400\)
\(144x^2+216x=144x^2-480x+400-81\)
\(114x^2+216=114x^2-480x+319\)
\(696x=319\)
\(\Rightarrow x=\frac{11}{24}\)
b) \(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\left(x-1\right)^2\left(x^2+2\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)
\(\Rightarrow x=1\)
c) \(x^5+x^4+x^3+x^2+x+1=0\)
\(\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow x=-1\)
Bài 2:
a) \(5x^3-7x^2-15x+21=0\)
\(\left(5x-7\right)\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)
\(\Rightarrow x=\frac{7}{5}\)
b) \(\left(x-3\right)^2=4x^2-20x+25\)
\(x^2-6x+9-25=4x^2-20x+25\)
\(x^2-6x+9=4x^2-20x+25-25\)
\(x^2-6x-16=4x^2-20x\)
\(x^2+14x-16=4x^2-4x^2\)
\(-3x^2+14x-16=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
c) \(\left(x-1\right)^2-5=\left(x+2\right)\left(x-2\right)-x\left(x-1\right)\)
\(x^2-2x=x-4\)
\(x^2-2x=x-4+4\)
\(x^2-2x=x-x\)
\(x^2-3x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
d) \(\left(2x-3\right)^3-\left(2x+3\right)\left(4x^2-1\right)=-24\)
\(-48x^2+56x-24=-24\)
\(-48x^2+56x=-24+24\)
\(-48x^2+56=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{6}\end{cases}}\)
mình ko chắc
\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)
Cảm ơn nhưng m lại ko hiểu làm sao ra đc như thế vậy. Bạn giải rõ ra đc ko