Bài 9 : Tìm x, biết :
a, ( x-2 ) ( x-3 ) + ( x-2 ) - 1 = 0
b, ( x+2 )^2 - 2x ( 2x + 3 ) = ( x+1 )^2
c, 6x^3 + x^2 = 2x
d, x^8 - x^5 + x^2 - x + 1 = 0
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28 tháng 8 2020
Ít thôi -..-
a) ( 3x + 2 )( 2x + 9 ) - ( x + 3 )( 6x + 1 ) = ( x + 1 )2 - ( x + 2 )( x - 2 )
<=> 6x2 + 31x + 18 - ( 6x2 + 19x + 3 ) = x2 + 2x + 1 - ( x2 - 4 )
<=> 6x2 + 31x + 18 - 6x2 - 19x - 3 = x2 + 2x + 1 - x2 + 4
<=> 12x + 15 = 2x + 5
<=> 12x - 2x = 5 - 15
<=> 10x = -10
<=> x = -1
b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )
<=> 2x2 - 5x - 12 + x2 - 7x + 10 = 3x2 - 17x + 20
<=> 3x2 - 12x - 2 = 3x2 - 17x + 20
<=> 3x2 - 12x - 3x2 + 17x = 20 + 2
<=> 5x = 22
<=> x = 22/5
c) ( x + 2 )3 - ( x - 2 )3 - 12x( x - 1 ) = -8
<=> x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 ) - 12x2 + 12x = -8
<=> x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
<=> 12x + 16 = -8
<=> 12x = -24
<=> x = -2
d) ( 3x - 1 )2 - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )2 = 16
<=> 9x2 - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x2 - 2x + 1 ) = 16
<=> 9x2 - 11x - 3 - x2 + 2x - 1 = 16
<=> 8x2 - 9x - 4 = 16
<=> 8x2 - 9x - 4 - 16 = 0
<=> 8x2 - 9x - 20 = 0
( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm
2 là nghiệm vô tỉ =) )
28 tháng 8 2020
a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)2 - (x + 2)(x - 2)
=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x2 + 2x + 1 - x(x - 2) - 2(x - 2)
=> 6x2 + 27x + 4x + 18 - 6x2 - x - 18x - 3 = x2 + 2x + 1 - x2 + 2x - 2x + 4
=> (6x2 - 6x2) + (27x + 4x - x - 18x) + (18 - 3) = (x2 - x2) + (2x + 2x - 2x) + (1 + 4)
=> 12x + 15 = 2x + 5
=> 12x + 15 - 2x - 5 = 0
=> 10x + 10 = 0
=> 10x = -10 => x = -1
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)
=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 = 3x2 - 12x - 5x + 20
=> (2x2 + x2) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x2 - 17x + 20
=> 3x2 - 12x - 2 = 3x2 - 17x + 20
=> 3x2 - 12x - 2 - 3x2 + 17x - 20 = 0
=> (3x2 - 3x2) + (-12x + 17x) + (-2 - 20) = 0
=> 5x - 22 = 0
=> 5x = 22 => x = 22/5
c) (x + 2)3 - (x - 2)3 - 12x(x - 1) = -8
=> x3 + 6x2 + 12x + 8 - (x3 - 6x2 + 12x - 8) - 12x2 + 12x = -8
=> x3 + 6x2 + 12x + 8 -x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
=> (x3 - x3) + (6x2 + 6x2 - 12x2) + (12x - 12x + 12x) + (8 + 8) = -8
=> 12x + 16 = -8
=> 12x = -24
=> x = -2
Còn bài cuối làm nốt
20 tháng 8 2021
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
20 tháng 7 2020
Tìm x
a) Ta có: \(3\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow3\left(x-1-4x^2+4x\right)+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow3\left(-4x^2+5x-1\right)+4\left(3x^2+11x+6\right)-38=0\)
\(\Leftrightarrow-12x^2+15x-3+12x^2+44x+24-38=0\)
\(\Leftrightarrow59x-17=0\)
\(\Leftrightarrow59x=17\)
hay \(x=\frac{17}{59}\)
Vậy: \(x=\frac{17}{59}\)
b) Ta có: \(5\left(2x+3\right)\left(x+2\right)-2\left(5x-4\right)\left(x-1\right)=75\)
\(\Leftrightarrow5\left(2x^2+4x+3x+6\right)-2\left(5x^2-5x-4x+4\right)-75=0\)
\(\Leftrightarrow5\left(2x^2+7x+6\right)-2\left(5x^2-9x+4\right)-75=0\)
\(\Leftrightarrow10x^2+35x+30-10x^2+18x-8-75=0\)
\(\Leftrightarrow53x-53=0\)
\(\Leftrightarrow53x=53\)
hay x=1
Vậy: x=1
c) Ta có: \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Leftrightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Leftrightarrow5x^2-3-5x^2-5x=0\)
\(\Leftrightarrow-3-5x=0\)
\(\Leftrightarrow-5x=-3\)
hay \(x=\frac{3}{5}\)
Vậy: \(x=\frac{3}{5}\)
d) Ta có: \(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow8x+16-5x^2-10x+4\left(x^2+x-2x-2\right)+2\left(x^2-4\right)=0\)
\(\Leftrightarrow-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: \(x\in\left\{0;6\right\}\)
20 tháng 7 2020
Tìm x
a) Ta có: \(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow16x^2-\left(16x^2-40x+25\right)-15=0\)
\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)
\(\Leftrightarrow40x-40=0\)
\(\Leftrightarrow40x=40\)
hay x=1
Vậy: x=1
b) Ta có: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(\Leftrightarrow4x^2+12x+9-4\left(x^2-1\right)-49=0\)
\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)
\(\Leftrightarrow12x-36=0\)
\(\Leftrightarrow12x=36\)
hay x=3
Vậy: x=3
d) Ta có: \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)
\(\Leftrightarrow12x-5=0\)
\(\Leftrightarrow12x=5\)
hay \(x=\frac{5}{12}\)
Vậy: \(x=\frac{5}{12}\)
e) Ta có: \(\left(x-5\right)^2-x\left(x-4\right)=9\)
\(\Leftrightarrow x^2-10x+25-x^2+4x-9=0\)
\(\Leftrightarrow-6x+16=0\)
\(\Leftrightarrow6x=16\)
hay \(x=\frac{8}{3}\)
Vậy: \(x=\frac{8}{3}\)
f) Ta có: \(\left(x-5\right)^2-\left(x-4\right)\left(1-x\right)=0\)
\(\Leftrightarrow x^2-10x+25-\left(x-x^2-4+4x\right)=0\)
\(\Leftrightarrow x^2-10x+25-x+x^2+4-4x=0\)
\(\Leftrightarrow2x^2-15x+29=0\)
\(\Leftrightarrow2\left(x^2-\frac{15}{2}x+\frac{29}{2}\right)=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{15}{4}+\frac{225}{16}+\frac{7}{16}=0\)
\(\Leftrightarrow\left(x-\frac{15}{4}\right)^2+\frac{7}{16}=0\)(vô lý)
Vậy: x∈∅
Bài 9 : Tìm x, biết :
a, (x - 2)(x - 3) + (x - 2) - 1 = 0
\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy x ={1; 3}
b, (x + 2)2 - 2x(2x + 3) = (x + 1)2
\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x
\(\Leftrightarrow6x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)
\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)
\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)