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1. (x + 2)(x2 - 2x + 4) - (x3 + 2x2) = 5
=> x(x2 - 2x + 4) + 2(x2 - 2x + 4) - x3 - 2x2 - 5 = 0
=> x3 - 2x2 + 4x + 2x2 - 4x + 8 - x3 - 2x2 - 5 = 0
=> (x3 - x3) + (-2x2 + 2x2 - 2x2) + (4x - 4x) + (8 - 5) = 0
=> -2x2 + 3 = 0
=> -2x2 = -3
=> x2 = 3/2
=> x = \(\pm\sqrt{\frac{3}{2}}\)
2. \(\left(x+5\right)^2-6=0\)
=> x2 + 10x + 25 - 6 = 0
=> x2 + 10x + 19 = 0
=> x vô nghiệm(do mình không để căn nên ghi vô nghiệm thôi nhá)
3. \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=2x\)
=> x(x2 - 3x + 9) + 3(x2 - 3x + 9) - x3 - 2x = 0
=> x3 - 3x2 + 9x + 3x2 - 9x + 27 - x3 - 2x = 0
=> (x3 - x3) + (-3x2 + 3x2) + (9x - 9x - 2x) + 27 = 0
=> -2x + 27 = 0
=> -2x = -27
=> x = 27/2
4. \(\left(x-2\right)^3-x^3+6x^2=7\)
=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
=> (x3 - x3) + (-6x2 + 6x2) + 12x - 8 = 7
=> 12x - 8 = 7
=> 12x = 15
=> x = 5/4
5. \(3\left(x-2\right)^2+9\left(x-1\right)-3\left(x^2+x-3\right)=12\)
=> 3x2 - 12x + 12 + 9x - 9 - 3x2 - 3x + 9 = 12
=> (3x2 - 3x2) + (-12x + 9x - 3x) + (12 - 9 + 9) = 12
=> -6x + 12 = 12
=> -6x = 0
=> x = 0
6. \(\left(4x+3\right)^2-\left(4x-3\right)^2-5x-2=0\)
=> 48x - 5x - 2 = 0
=> 43x - 2 = 0
=> 43x = 2
=> x = 2/43
Còn bài cuối tự làm :>
Anh Sang làm cầu kì quá ;-;
1. ( x + 2 )( x2 - 2x + 4 ) - ( x3 + 2x2 ) = 5
<=> x3 + 8 - x3 - 2x2 = 5
<=> 8 - 2x2 = 5
<=> 2x2 = 3
<=> x2 = 3/2
<=> \(x^2=\left(\pm\sqrt{\frac{3}{2}}\right)^2\)
<=> \(x=\pm\sqrt{\frac{3}{2}}\)
2. ( x + 5 )2 - 6 = 0
<=> ( x + 5 )2 - ( √6 )2 = 0
<=> ( x + 5 - √6 )( x + 5 + √6 ) = 0
<=> \(\orbr{\begin{cases}x+5-\sqrt{6}=0\\x+5+\sqrt{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-5\\x=-\sqrt{6}-5\end{cases}}\)
3. ( x + 3 )( x2 - 3x + 9 ) - x3 = 2x
<=> x3 + 27 - x3 = 2x
<=> 27 = 2x
<=> x = 27/2
4. ( x - 2 )3 - x3 + 6x2 = 7
<=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
<=> 12x - 8 = 7
<=> 12x = 15
<=> x = 15/12 = 5/4
5. 3( x - 2 )2 + 9( x - 1 ) - 3( x2 + x - 3 ) = 12
<=> 3( x2 - 4x + 4 ) + 9x - 9 - 3x2 - 3x + 9 = 12
<=> 3x2 - 12x + 12 + 6x - 3x2 = 12
<=> -6x + 12 = 12
<=> -6x = 0
<=> x = 0
6. ( 4x + 3 )2 - ( 4x - 3 )2 - 5x - 2 = 0
<=> 16x2 + 24x + 9 - ( 16x2 - 24x + 9 ) - 5x - 2 = 0
<=> 16x2 + 24x + 9 - 16x2 + 24x - 9 - 5x - 2 = 0
<=> 43x - 2 = 0
<=> 43x = 2
<=> x = 2/43
7, ( 4x + 7 )( 2 - 3x ) - ( 6x + 2 )( 5 - 2x ) = 0
<=> -12x2 - 13x + 14 - ( -12x2 + 26x + 10 ) = 0
<=> -12x2 - 13x + 14 + 12x2 - 26x - 10 = 0
<=> -39x + 4 = 0
<=> -39x = -4
<=> x = 4/39
Bài 1:
a) (3x-2).(4x+5)-6x.(2x-1) = 12x^2 +15x - 8x -10 - 12x^2 + 6x = 13x - 10
b) (2x-5)^2 - 4.(x+3).(x-3) = 4x^2 - 20x + 25 - 4x^2 + 12x -12x + 36 = -20x + 61
Bài 2:
a)(2x-1)^2-(x+3)^2 = 0
<=> (2x-1-x-3).(2x-1+x+3) =0
<=>(x-4).(3x+2) = 0
<=> x-4 = 0 hoặc 3x+2=0
*x-4=0 => x=4
*3x+2 = 0 => 3x=-2 => x=-2/3
b)x^2(x-3)+12-4x=0 <=> x^2(x-3) - 4(x-3) =0 <=> (x-3).(x-2)(x+2) <=> x-3=0 hoặc x-2=0 hoặc x+2 =0
*x-3=0 => x=3
*x-2=0 =>x=2
*x+2=0 =>x=-2
c) 6x^3 -24x =0 <=> 6x(x^2 -4)=0 <=> 6x(x-2)(x+2)=0 <=> x=0 hoặc x-2 =0 hoặc x+2=0 <=> x=0 hoặc x=2 hoặc x=-2
Đề bài là giải các phương trình nha :Đ
\(b,\left(2x+1\right)^2=9\)
\(\left(2x+1\right)^2=3^2\)
\(\Rightarrow\orbr{\begin{cases}2x+1=3\\2x+1=-3\end{cases}\Rightarrow\orbr{\begin{cases}2x=2\\2x=-4\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)
\(c,x^3+5x^2-4x-20=0\)
\(x^2\left(x+5\right)-4\left(x+5\right)=0\)
\(\left(x^2-4\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-4=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=4\\x=5\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases};x=5}\)
ko phải mk lười đâu , cái này bn làm nó mới có ý nghĩa , cố gắng nốt nha !
Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)
Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
1. <=> \(\left(3x+2\right)^3-\left(\left(3x\right)^3+2^3\right)=0\)
<=> \(\left(\left(3x\right)^3+2^3+3\left(3x+2\right).3x.2\right)-\left(\left(3x\right)^3+2^3\right)=0\)
<=>3 (3x + 2) . 3x.2 = 0
<=> (3x + 2 ) . x = 0
<=> x = -2/3 hoặc x = 0
2. Tương tự
1
\(\left(3x+2\right)^3-\left[\left(3x\right)^3+2^3\right]=0\)
\(\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot2+3\cdot3x\cdot2^2+2^3-\left(3x\right)^3-2^3=0\)
\(54x^2+36x=0\)
\(18x\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{-2}{3}\end{cases}}\)
2
\(\left(2x+1\right)^3-\left[\left(2x\right)^3-1^3\right]=0\)
\(\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3-\left(2x\right)^3-1^3=0\)
\(12x^2+6x=0\)
\(6x\left(2x+1\right)=0\)
\(\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)
f: Ta có: \(x\left(2x-9\right)-4x+18=0\)
\(\Leftrightarrow\left(2x-9\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=2\end{matrix}\right.\)
g: Ta có: \(4x\left(x-1000\right)-x+1000=0\)
\(\Leftrightarrow\left(x-1000\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1000\\x=\dfrac{1}{4}\end{matrix}\right.\)
f. x(2x - 9) - 4x + 18 = 0
<=> x(2x - 9) - 2(2x - 9) = 0
<=> (x - 2)(2x - 9) = 0
<=> \(\left[{}\begin{matrix}x-2=0\\2x-9=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\x=\dfrac{9}{2}\end{matrix}\right.\)
g. 4x(x - 1000) - x + 1000 = 0
<=> 4x(x - 1000) - (x - 1000) = 0
<=> (4x - 1)(x - 1000) = 0
<=> \(\left[{}\begin{matrix}4x-1=0\\x-1000=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=1000\end{matrix}\right.\)
h. 2x(x - 4) - 6x2(-x + 4) = 0
<=> 2x(x - 4) + 6x2(x - 4) = 0
<=> (2x + 6x2)(x - 4) = 0
<=> 2x(1 + 3x)(x - 4) = 0
<=> \(\left[{}\begin{matrix}2x=0\\1+3x=0\\x-4=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{3}\\x=4\end{matrix}\right.\)
i. 2x(x - 3) + x2 - 9 = 0
<=> 2x(x - 3) + (x - 3)(x + 3) = 0
<=> (2x + x + 3)(x - 3) = 0
<=> (3x + 3)(x + 3) = 0
<=> \(\left[{}\begin{matrix}3x+3=0\\x+3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
j. 9x - 6x2 + x3 = 0
<=> x(9 - 6x + x2) = 0
<=> x(3 - x)2 = 0
<=> \(\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)