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Mk lm câu b bài 2 há!
b, ( 8x - 3 )( 3x + 2 ) - ( 4x + 7 )( x + 4 ) = ( 2x +1 )( 5x - 1) =- 33
Pt <=> 3x ( 8x - 3 ) + 2( 8x- 33) - ( x ( 4x + 7) ) + ( 2x + 1) - 5x ( 2x + 1) + 33 = 0
<=> 24x2 - 9x + 16x - 6 - ( 4x2 + 7x + 16x + 28) + 2x + 1 - 10x2 - 5x + 33 = 0
<=> 24x2 - 9x + 16x - 6 - 4x2 - 7x - 16x - 28 + 2x + 1 - 10x2 - 19x = 0 <=> x ( 10x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\10x-19=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{19}{10}\end{cases}}\)
^^ Ok con tê tê!
Bài 2: Vì: 2m - 2n = 256 nên m> n
Đặt m - n = d ( d > 0 )
Ta có : 2m - 2n = 2n ( 2d - 1 ) = 256 = 28.1
=> 2n = 28 và 2d - 1 = 1
=> n = 8 và d = 1
=> m = 1 + 8 = 10
Vậy n = 8 ; m = 9
Bài 1:
a/ \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)
\(=\left(4x-4\right)\left(x+5\right)-\left(x^2+5x+2x+10\right)-\left(3x-3\right)\left(x+2\right)\)
\(=4x^2+20x-4x-20-x^2-5x-2x-10-3x^2-3x-6x-6\)
\(=-36\)
b/ \(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\left(x^{3n}+y^{3n}\right)\)
\(=\left(x^{3n}+x^{2n}y^n+x^ny^{2n}-x^{2n}y^n-x^ny^{2n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)
\(=x^{6n}+x^{5n}y^n+x^{4n}y^{2n}-x^{5n}y^n-x^{4n}y^{2n}-x^{3n}y^{3n}+x^{3n}y^{3n}+x^{2n}y^{4n}+x^ny^{5n}-x^{2n}y^{4n}-x^ny^{5n}-y^{6n}\)
\(=x^{6n}-y^{6n}\)
Bài 2:
a/ \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-5x-2x+10=3x^2-5x-12x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
\(\Leftrightarrow3x^2-3x^2-12x+17x=20+2\)
\(5x=22\Rightarrow x=\dfrac{22}{5}\)
Vậy...............
b/ Tương tự!
Huyền Anh Kute câu a đúng r`
b, \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1-33\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-34\)
\(\Leftrightarrow20x^2-16x-10x^2-3x=-34+34\)
\(\Leftrightarrow10x^2-19x=0\)
\(\Leftrightarrow x\left(10x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\10x-19=0\Rightarrow x=\dfrac{19}{10}\end{matrix}\right.\)
Vậy........................
Câu 1:
a: =>(x-3)(x+3)(2x+1)=0
hay \(x\in\left\{3;-3;-\dfrac{1}{2}\right\}\)
b: \(\Leftrightarrow x^2-2x+1+6=0\)
\(\Rightarrow\left(x-1\right)^2+6=0\)(vô lý)