Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a; (\(x+1\)).(\(x+2\)) - (\(x-1\)).(\(x-5\)) = 0
\(x^2\) + 2\(x\) + \(x+2\) - \(x^2\) + 5\(x\) + \(x\) - 5 = 0
(\(x^2\) - \(x^2\)) + (2\(x\) + \(x+5x+x\))- (5 -2) = 0
0 + (3\(x\) + 5\(x\) + \(x\)) + 0 - 3 = 0
8\(x\) + \(x\) - 3 = 0
9\(x\) = 3
\(x=\dfrac{3}{9}\)
Vậy \(x=\dfrac{1}{3}\)
b; (2\(x\) - 1)2 + 4.(5 - \(x\)) = 15
4\(x^2\) - 4\(x\) + 1 + 20 - 4\(x\) = 15
4\(x^2\) - (4\(x\) + 4\(x\)) + (1 + 20 - 15) = 0
4\(x^2\) - 8\(x\) + 6 = 0
4.(\(x^2\) - 2\(x\) + 1) + 2 = 0
4(\(x-1\))2 + 2 = 0
Vì 4.(\(x-1\))2 ≥ 0 ⇒ 4.(\(x-1\))2 + 2 ≥ 3 > 0 (\(\forall x\))
Vậy không có giá trị nào của \(x\) thỏa mãn đề bài
Kết luận \(x\) \(\in\) \(\varnothing\)
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a) \(5\left(x+7\right)-12x=15\)
\(5x+35-12x=15\)
\(-7x=15-35\)
\(-7x=-20\)
\(x=\frac{20}{7}\)
vay \(x=\frac{20}{7}\)
b) \(x^2-25-\left(x+5\right)=0\)
\(x^2-5^2-\left(x+5\right)=0\)
\(\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(x-5-1\right)=0\)
\(\left(x+5\right)\left(x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)
vay \(\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)
c) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\left(2x-1\right)\left(2x-1\right)-\left(\left(2x\right)^2-1^2\right)=0\)
\(\left(2x-1\right)\left(2x-1\right)-\left(2x-1\right)\left(2x+1\right)=0\)
\(\left(2x-1\right)\left(2x-1-2x-1\right)=0\)
\(-2.\left(2x-1\right)=0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow x=\frac{1}{2}\)
vay \(x=\frac{1}{2}\)
d) \(x^2.\left(x^2+4\right)-x^2-4=0\)
\(x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-1=0\\x^2+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=1\\x^2=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=1hoacx=-1\\kotontai\end{cases}}\)
vay \(x=1\)hoac \(x=-1\)
a) Ta có : 6x(3x + 5) - 2x(9x - 2) + (17 - x)(x - 1) + x(x - 18) = 0
<=> 18x2 + 30x - 18x2 + 4x + 17x - 17 - x2 + x + x2 - 18x = 0
<=> 34x - 17 = 0
<=> 34x = 17
=> x = 2
b) ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0
<=> ( 2x - 3 ) + ( 2x - 3 )( x - 1 ) = 0
<=> ( 2x - 3 )( 1 + x - 1 ) = 0
<=> x( 2x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)
Vậy .....
a, 25x^2 - 1 - (5x -1)(x+2)=0
=> (5x)^2 - 1 + (5x-1)(x+2) = 0
=> (5x-1)(5x+1) + (5x-1)(x+2) = 0
=> (5x-1)(5x+1+x+2) = 0
=> (5x-1)(6x+3) = 0
=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)
Trả lời:
a, \(\left(3x+1\right)\left(x-3\right)-x\left(3x-14\right)=15\)
\(\Leftrightarrow3x^2-9x+x-3-3x^2+14x=15\)
\(\Leftrightarrow6x-3=15\)
\(\Leftrightarrow6x=18\)
\(\Leftrightarrow x=3\)
Vậy x = 3 là nghiệm của pt.
b, \(\left(x-3\right)^2=9-x^2\)
\(\Leftrightarrow\left(x-3\right)^2-9+x^2=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-3+x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right).2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=0\end{cases}}}\)
Vậy x = 3; x = 0 là nghiệm của pt.
c, \(\left(2x-\frac{1}{2}\right)^2-\left(1-2x\right)^2=2\)
\(\Leftrightarrow4x^2-2x+\frac{1}{4}-\left(1-4x+4x^2\right)=2\)
\(\Leftrightarrow4x^2-2x+\frac{1}{4}-1+4x-4x^2=2\)
\(\Leftrightarrow2x-\frac{3}{4}=2\)
\(\Leftrightarrow2x=\frac{11}{4}\)
\(\Leftrightarrow x=\frac{11}{8}\)
Vậy x = 11/8 là nghiệm của pt.
d, \(4x^2+4x-3=0\)
\(\Leftrightarrow4x^2-2x+6x-3=0\)
\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
Vậy x = 1/2; x = - 3/2 là nghiệm của pt.
a) ( 2x + 3 )^2 - 4( x - 1 )( x + 1 ) = 49
=>4x2+12x+9-4x2+4=49
=>12x+13=49
=>12x=36
=>x=3
b) 16x^2 - ( 4x - 5 )^2 = 15
=>16x2-16x2+40x-25=15
=>40x-25=15
=>40x=40
=>x=1
c) ( 2x + 1 )^2 - ( x - 1)^2 = 0
=>4x2+4x+1-x2+2x-1=0
=>3x2+6x=0
=>3x(x+2)=0
=>3x=0 hoặc x+2=0
=>x=0 hoặc x=-2
a) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ =>4x^2+12x+9-4x^2+4=49\\=>12x+13=49\\ =>12x=36\\ =>x=3\)
b) \(16x^2-\left(4x-5\right)^2=15\\ =>16x^2-16x^2+40x-25=15\\ =>40x-25=15\\ =>40x=40\\ =>x=1\)
c) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\\ =>4x^2+4x+1-x^2+2x-1=0\\ =>3x^2+6x=0\\ =>3x\left(x+2\right)=0\\ =>\left[\frac{3x=0}{x+2=0}\right]=>\left[\frac{x=0}{x=-2}\right]\)
\(b;\left(x+1\right)^2=x+1\)
\(\Rightarrow\left(x+1\right)^2-x-1=0\)
\(\Rightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+1-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)
a; \(x\left(x+1\right)\) - (\(x+1\))2 = 5
(\(x-x-1\))(\(x+1\))= 5
(0 - 1).(\(x+1\)) = 5
-1.(\(x+1\)) = 5
\(x+1\) = -5
\(x=-5-1\)
\(x=-6\)
Vậy \(x=-6\)
b; \(x^2\) - 4\(x=0\)
\(x\).(\(x-4\)) = 0
\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x\) \(\in\) {0; 4}
khó vậy em ko biết