Tìm x , biết
a, 5x3 - 125 = 0
b, x2 - 6x-21 =0
c, 6x + 13x + 5 = 0
d, x . ( x-7) -14x + 28 =0
. Tìm x, biết:
a) 4x2 – 9 = 0
b) (x + 5)2 – (x – 1)2= 0
c) x2 – 6x – 7 = 0
d) (x + 1)2 – (2x - 1)2 = 0
a)4x2-9=0
⇔ (2x-3)(2x+3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b)(x+5)2-(x-1)2=0
⇔ (x+5-x+1)(x+5+x-1)=0
⇔ 12(x+2)=0
⇔ x=-2
c)x2-6x-7=0
⇔ x2-7x+x-7=0
⇔ x(x-7)+(x-7)=0
⇔ (x-7)(x+1)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
d)(x+1)2-(2x-1)2=0
⇔ (x+1-2x+1)(x+1+2x-1)=0
⇔3x(2-x)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
a, 4x2 - 9 = 0
<=> 4x2 = 9
<=> x2 = \(\dfrac{9}{4}\) => x = \(\sqrt{\dfrac{9}{4}}\)
b, (x + 5 )2 - ( x - 1 )2 = 0
<=> ( x+5-x+1 )(x+5+x-1) = 0
<=> 6(2x+4) = 0
<=> 12x+24=0
<=> 12x = -24
<=> x = -2
c, x2-6x-7=0
<=> x2+x-7x-7=0
<=> x(x+1)-7(x+1)=0
<=> (x-7)(x+1)=0
=> x+7=0 hoặc x+1=0
+ x-7=0 => x=7
+ x+1=0 => x=-1
d, \(\left(x+1\right)^2-\left(2x-1\right)^2=0\)
<=> \(\left(x+1-2x+1\right)\left(x+1+2x-1\right)=0\)
<=> (-x+2).3x=0
=> x=0 hoặc (-x+2).3=0
+ (-x+2).3=0 => -3x+6=0 => x=-2
b) (x +5)2 -(x -1)2=0
<=> [(x +5) -(x -1)][(x +5) +(x -1)]=0
<=> (x +5 -x +1)(x +5 +x -1)=0
<=> 6(2x+4)=0 <=>12(x +2)=0
=> x +2=0=> x=-2
vậy x= -2
c) x2 -6x -7=0
<=> x2 -7x +x -7=0
<=> (x2 +x)( -7x -7)=0
<=> x(x +1).-7(x +1)=0
<=> (x +1)(x -7)=0
<=> \(\left\{{}\begin{matrix}x+1=0\\x-7=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)
Vậy S={-1; 7}
d) (x +1)2 -(2x -1)2=0
<=> [(x -1)-(2x -1)][(x -1)+(2x -1)]=0
<=> (x -1 -2x +1)(x -1 +2x -1)=0
<=> (x -2x)(3x -2)<=> -x(3x -2)=0
<=> \(\left\{{}\begin{matrix}-x=0\\3x-2=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy S={0; \(\dfrac{2}{3}\)}
Giải các phương trình sau:
a) 4 − 5 x = 5 − 6 x ; b) 3 x + 2 − 7 x + 1 = 0 ;
c) x 2 − 2 x − 3 + x + 1 = 0 ; d) 1 4 x − 5 = 3 x + 1
a) Trường hợp 1. Xét 4 - 5x = 5 - 6x.
Tìm được x = 1.
Tìm x biết :
a) 6x2 + 5x - 6 = 0
b) 6x2 - 13x + 6 = 0
c) 10x2 - 13x - 3 =0
d) 20x2 + 19x - 3 = 0
e) 3x2 -x + 6 = 0
a)\(6x^2+5x-6=0\)
\(\Leftrightarrow6x^2-4x+9x-6=0\)
\(\Leftrightarrow2x\left(3x-2\right)+3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
b)\(6x^2-13x+6=0\)
\(\Leftrightarrow6x^2-4x-9x+6=0\)
\(\Leftrightarrow2x\left(3x-2\right)-3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
c)\(10x^2-13x-3=0\)
\(\Leftrightarrow10x^2-15x+2x-3=0\)
\(\Leftrightarrow5x\left(2x-3\right)+\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\5x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{5}\end{array}\right.\)
d)\(20x^2+19x-3=0\)
\(\Delta=19^2-\left(-4\left(20.3\right)\right)=601\)
\(\Rightarrow x_{1,2}=\frac{-19\pm\sqrt{601}}{40}\)
e)\(3x^2-x+6=0\)
\(\Delta=\left(-1\right)^2-4\left(3.6\right)=-71< 0\)
Suy ra vô nghiệm
Tìm x, biết
a) 4(x-2)2=4
b) 5(x2-6x+9)=5
c) 4x2+4x+1=0
d) 9x2+6x+1=2
a)
`4(x-2)^2 =4`
`<=>(x-2)^2 =1`
`<=>x-2=1` hoặc `x-2=-1`
`<=>x=3` hoặc `x=1`
b)
`5(x^2 -6x+9)=5`
`<=>(x-3)^2 =1`
`<=>x-3=1`hoặc `x-3=-1`
`<=>x=4` hoặc `x=2`
c)
`4x^2 +4x+1=0`
`<=>(2x+1)^2 =0`
`<=>2x+1=0`
`<=>x=-1/2`
d)
`9x^2 +6x+1=2`
`<=>(3x+1)^2 =2`
\(< =>\left[{}\begin{matrix}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{\sqrt{2}-1}{3}\\x=\dfrac{-\sqrt{2}-1}{3}\end{matrix}\right.\)
tìm x biết
a,2x^2-6x+4=0
b,5x^2-10x+4=0
c,x^2+7x+12=0
d,13x^2+15x-10=0
g,7x^2-4x-1=0
Bài 1: Tìm số tự nhiên x, biết:
a,36 : [ 7(x - 3) + 4 ] = 24 :23
b,[(6x - 39) : 3 ] . 28 = 5628
c,(2x -7) - (x + 135)=0
d,24 .125 + 52 . 25
e,17 . 27 +17 . 25 + 17 . 48
g,122 + (37 - 3x) = 0
h,(14 - 3x) + (6+x) = 0
Giải các phương trình sau:
a) \(6x + 4 = 0\);
b) \( - 14x - 28 = 0\);
c) \(\frac{1}{3}x - 5 = 0\);
d) \(3y - 1 = - y + 19\);
e) \( - 2\left( {z + 3} \right) - 5 = z + 4\);
g) \(3\left( {t - 10} \right) = 7\left( {t - 10} \right)\)
a)
\(\begin{array}{l}6x + 4 = 0\\\,\,\,\,\,\,\,\,6x = - 4\\\,\,\,\,\,\,\,\,\,\,x = \left( { - 4} \right):6\\\,\,\,\,\,\,\,\,\,\,x = - \frac{2}{3}.\end{array}\)
Vậy phương trình có nghiệm \(x = - \frac{2}{3}.\)
b)
\(\begin{array}{l} - 14x - 28 = 0\\\,\,\,\,\,\,\,\,\, - 14x = 28\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 28:\left( { - 14} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = - 2\end{array}\)
Vậy phương trình có nghiệm \(x = - 2.\)
c)
\(\begin{array}{l}\frac{1}{3}x - 5 = 0\\\,\,\,\,\,\,\,\frac{1}{3}x = 5\\\,\,\,\,\,\,\,\,\,\,\,x = 5:\frac{1}{3}\\\,\,\,\,\,\,\,\,\,\,\,x = 15.\end{array}\)
Vậy phương trình có nghiệm \(x = 15\).
d)
\(\begin{array}{l}\,3y - 1 = - y + 19\\3y + y = 19 + 1\\\,\,\,\,\,\,\,4y = 20\\\,\,\,\,\,\,\,\,\,\,y = 20:5\\\,\,\,\,\,\,\,\,\,\,y = 4.\end{array}\)
Vậy phương trình có nghiệm \(y = 4\).
e)
\(\begin{array}{l} - 2\left( {z + 3} \right) - 5 = z + 4\\\,\,\, - 2z - 6 - 5 = z + 4\\\,\,\,\,\,\,\,\, - 2z - 11 = z + 4\\\,\,\,\,\,\,\,\,\,\, - 2z - z = 4 + 11\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3z = 15\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z = 15:\left( { - 3} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z = - 5.\end{array}\)
Vậy phương trình có nghiệm \(z = - 5\).
g)
\(\begin{array}{l}3\left( {t - 10} \right) = 7\left( {t - 10} \right)\\\,\,\,\,3t - 30 = 7t - 70\\\,\,\,\,\,3t - 7t = - 70 + 30\\\,\,\,\,\,\,\,\,\,\, - 4t = - 40\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t = \left( { - 40} \right):\left( { - 4} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t = 10.\end{array}\)
Vậy phương trình có nghiệm \(t = 10\).
Tìm x biết rằng:
a) ( x 2 + 2x + 4)(2 - x) + x(x - 3)(x + 4) - x 2 + 24 = 0;
b) x 2 + 3 ( 5 − 6 x ) + ( 12 x − 2 ) x 4 + 3 = 0 .
Tìm x :
a) x2 – 15x + 14 = 0
b) x2 + 9x + 14 = 0
c) 6x2 – 11x + 5 = 0
d) 6x2 + 13x + 5 = 0
e) 10x2 + 13x +3 = 0
a, \(x^2-x-14x+14=0\)
\(=>x\left(x-1\right)-14\left(x-1\right)=0\)
\(=>\left(x-14\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-14=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=14\\x=1\end{matrix}\right.\)
b, \(x^2+2x+7x+14=0\)
\(=>x\left(x+2\right)+7\left(x+2\right)=0\)
\(=>\left(x+7\right)\left(x+2\right)=0\)
\(< =>\left\{{}\begin{matrix}x+7=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=-2\end{matrix}\right.\)
c, \(6x^2-6x-5x+5=0\)
\(=>6x\left(x-1\right)-5\left(x-1\right)=0\)
\(=>\left(6x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}\\x=1\end{matrix}\right.\)
d, \(6x^2+3x+10x+5=0\)
\(=>3x\left(2x+1\right)+5\left(2x+1\right)=0\)
\(=>\left(3x+5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
e, \(10x^2+10x+3x+3=0\)
\(=>10x\left(x+1\right)+3\left(x+1\right)=0\)
\(=>\left(10x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x+3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{10}\\x=-1\end{matrix}\right.\)
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