Tìm x biết
b) ( 2x + 3)^2 – (2x + 1)(2x – 1) = 22
c) ( 4x + 3)( 4x – 3) – (4x -5)^2 = 16 d) x^3 – 9x^2+ 27x – 27 = - 8
e) (x + 1)^3 – x2 (x + 3) = 2 ; f) (x – 2)^3 – x( x – 1)(x + 1) + 6x2 =5
Tìm x biết
a) (x-3)^2 -4=0
b) ( 2x+3)^2 - (2x+1)(2x-1) =22
c) (4x+3)(4x-3) -(4x-5)^2 =16
d) x^3 -9x^2 +27x-27 =-8
e) (x+1)^3 - x^2(x+3) =2
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=0+4\)
\(\left(x-3\right)^2=4\)
\(\left(x-3\right)^2=\pm4\)
\(\left(x-3\right)^2=\pm2^2\)
\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(x=1\)
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
\(16x^2-9-16x^2+40x-25=16\)
\(-34+40x=16\)
\(40x=16+34\)
\(40x=50\)
\(x=\frac{50}{40}=\frac{5}{4}\)
d) \(x^3-9x^2+27x-27=-8\)
\(x^3-9x^2+27x-27+8=0\)
\(x^3-9x^2+27x-19=0\)
\(\left(x^2-8x+19\right)\left(x-1\right)=0\)
Vì \(\left(x^2-8x+19\right)>0\) nên:
\(x-1=0\)
\(x=1\)
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)
\(3x+1=2\)
\(3x=2-1\)
\(3x=1\)
\(x=\frac{1}{3}\)
Tìm x biết
a) (x-3)^2 -4=0
b) ( 2x+3)^2 - (2x+1)(2x-1) =22
c) (4x+3)(4x-3) -(4x-5)^2 =16
d) x^3 -9x^2 +27x-27 =-8
e) (x+1)^3 - x^2(x+3) =2
b) ( 2x+3)^2 - (2x+1)(2x-1) =22
=> 4x2+12x+9-4x2+1=22
=> 12x=12
=>x=1
c) (4x+3)(4x-3) -(4x-5)^2 =16
=>16x2-9-16x2+40x-25=16
=>40x=50
=>x=4/5
a)\(\left(x-13\right)^2-4=0\\\left(x-13\right)^2=4\\ \left(x-13\right)^2=2^2\\ \Rightarrow\left\{{}\begin{matrix}x-13=2\\x-13=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}15\\-11\end{matrix}\right.\)
vậy...
d) x^3 -9x^2 +27x-27 =-8
=>(x-3)3=-8
=>x-3=-2
=>x=1
e) (x+1)^3 - x^2(x+3) =2
=>x3+3x2+3x+1-x3-3x2=2
=>3x=1
=>x=1/3
Bài 4: Tìm x biết
a) (x-3) mũ 2 -4=0
b) (2x+3) mũ 2 - (2x+1)(2x-1)=22
c) (4x+3)(4x-3) - (4x-5) mũ 2=16
d) x mũ 3 - 9x mũ 2 + 27x - 27= -8
e) (x+1) mũ 3 - x mũ 2 nhân (x+3)=2
f) (x-2) mũ 3 - x(x-1)(x+1) + 6x mũ 2=5
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 - 22 = 0
<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0
<=> ( x - 5 )( x - 1 ) = 0
<=> x = 5 hoặc x = 1
b( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22
<=> 4x2 + 12x + 9 - ( 4x2 - 1 ) = 22
<=> 4x2 + 12x + 9 - 4x2 + 1 = 22
<=> 12x + 10 = 22
<=> 12x = 12
<=> x = 1
c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )2 = 16
<=> 16x2 - 9 - ( 16x2 - 40x + 25 ) = 16
<=> 16x2 - 9 - 16x2 + 40x - 25 = 16
<=> 40x - 34 = 16
<=> 40x = 50
<=> x = 50/40 = 5/4
d) x3 - 9x2 + 27x - 27 = -8
<=> ( x - 3 )3 = -8
<=> ( x - 3 )3 = (-2)3
<=> x - 3 = -2
<=> x = 1
e) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
f) ( x - 2 )3 - x( x - 1 )( x + 1 ) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5
<=> x3 + 12x - 8 - x3 + x = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
a) \(\left(x-3\right)^2-4=0\)
=> \(\left(x-3\right)^2-2^2=0\)
=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)
=> \(\left(x-5\right)\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)
=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)
=> \(4x^2+12x+9-4x^2+1=22\)
=> \(12x+9+1=22\)
=> \(12x+10=22\)
=> 12x = 12
=> x = 1
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)
=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)
=> \(16x^2-9-16x^2+40x-25=16\)
=> \(-9+40x-25=16\)
=> \(40x=16+25-\left(-9\right)=16+25+9=50\)
=> x = 50/40 = 5/4
d) \(x^3-9x^2+27x-27=-8\)
=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)
=> \(\left(x-3\right)^3=-8\)
=> \(\left(x-3\right)^3=\left(-2\right)^3\)
=> x - 3 = -2 => x = 1
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)
=> \(3x+1=2\)
=> \(3x=1\)=> x = 1/3
f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)
=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)
=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)
=> \(\left(12x+x\right)-8=5\)
=> 13x = 13
=> x = 1
a) (x+3)^2-4=0
=>(x+3)^2 = 4
=>(x+3)^2 = 2^2 = (-2)^2
=>x+3 = 2 hoặc -2
=> x= -1 hoặc -5
Tìm x, biết:
a) x3 - 9x2 + 27x - 27 = -8
b) 64x3 + 48x2 + 12x + 1 = 27
c) (2x - 1)3 - 4x2. (2x - 3) = 5
d) (x + 4)3 - x2. (x + 12) = 16
a) x3 - 9x2 + 27x - 27 = -8
<=> x3 - 3x2.3 + 3x.32 - 33 = -8
<=> (x - 3)3 = -23
<=> x - 3 = -2
<=> x = 1 (T/m)
Vậy x = 1.
b) 64x3 + 48x2 + 12x + 1 = 27
<=> (4x)3 + 3.(4x)2.1 + 3.4x.12 + 13 = 27
<=> (4x + 1)3 = 33
<=> 4x + 1 = 3
<=> 4x = 2
<=> x = \(\frac{1}{2}\)(T/m)
Vậy x = \(\frac{1}{2}\).
c) (2x - 1)3 - 4x2.(2x - 3) = 5
<=> (8x3 - 12x2 + 6x - 1) - (8x3 - 12x2) = 5
<=> 8x3 - 12x2 + 6x - 1 - 8x3 + 12x2 = 5
<=> 6x - 1 = 5
<=> 6x = 6
<=> x = 1 (T/m)
Vậy x = 1.
Tìm x, biết:
a) x3 - 9x2 + 27x - 27 = -8
b) 64x3 + 48x2 + 12x + 1 = 27
c) (2x - 1)3 - 4x2. (2x - 3) = 5
d) (x + 4)3 - x2. (x + 12) = 16
\(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x^3-27\right)-\left(9x^2-27x\right)=\left(x-3\right)\left(x^2+3x+9\right)-9x\left(x-3\right)=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3=-8=\left(-2\right)^3\Rightarrow x=\left(-2\right)+3=1\)
\(64x^3+48x^2+12x+1=\left(64x^3+1\right)+\left(48x^2+12x\right)=\left(4x+1\right)\left(16x^2-4x+1\right)+12x\left(4x+1\right)=\left(4x+1\right)\left(16x^2+8x+1\right)=\left(4x+1\right)^3=27\Rightarrow4x=2\Leftrightarrow x=\frac{1}{2}\)
c) \(\left(2x-1\right)^3-4x^2.\left(2x-3\right)=5\)
\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(8x^3-12x^2\right)=5\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)
\(\Leftrightarrow6x-1=5\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\)
d) \(\left(x+4\right)^3-x^2.\left(x+12\right)=16\)
\(\Leftrightarrow\left(x^3+12x^2+48x+64\right)-\left(x^3+12x^2\right)=16\)
\(\Leftrightarrow x^3+12x^2+48x+64-x^3-12x^2=16\)
\(\Leftrightarrow48x+64=16\)
\(\Leftrightarrow48x=-48\)
\(\Leftrightarrow x=-1\)
#vì câu a,b có người làm rồi nên mình chỉ làm c,d thôi nhé ! :)
Học Tốt !!
Bài 5: Tìm nghiệm của các đa thức sau: Dạng 1: a) 4x + 9 b) -5x + 6 c) 7 – 2x d) 2x + 5 Dạng 2: a) ( x+ 5 ) ( x – 3) b) ( 2x – 6) ( x – 3) c) ( x – 2) ( 4x + 10 ) Dạng 3: a) x2 -2x b) x2 – 3x c) 3x2 – 4x d) ( 2x- 1)2 Dạng 4: a) x2 – 1 b) x2 – 9 c)– x 2 + 25 d) x2 - 2 e) 4x2 + 5 f) –x 2 – 16 g) - 4x4 – 25 Dạng 5: a) 2x2 – 5x + 3 b) 4x2 + 6x – 1 c) 2x2 + x – 1 d) 3x2 + 2x – 1
Giúp mình với
a)3y-2y=2y-3
b)3-4x+24+6x=x+27+3x
c)5-(6-x)=4(3-2x)
d)4(x+3)=-7x+17
e)11x+42-2x=100-9x-22
g)2x+3/3 =5-4x/2
f)5x+3/12 = 1+2x/9
h)7x-1/6 = 16-x/5
i)x-3/5 = 6-1-2x/3
k)3x-2/6 - 5=3-2(x+7)/4
3x-7/2 + x+1/3=-16
x-x+1/3 = 2x+1/5
2x-1/3 - 5x+2/7
a, 3y-2y=2y-3
3y-2y-2y=3
-y=3
y=-3
b, 3-4x+24+6x=x+27+3x
-4x+6x-x-3x =27-3-24
-2x =0
x =0
c, 5-(6-x)=4.(3-2x)
5-6+x =12-8x
x+8x =12+6-5
9x =13
x =13/9
d, 4.(x+3)=-7x+17
4x+12 =-7x+17
4x+7x =17-12
11x =5
x =5/11
A) \(3y-2y=2y-3\) \(\Leftrightarrow y-2y=-3\) \(\Leftrightarrow y=3\)
B) \(3-4x+24+6x=x+27-3x\)\(\Leftrightarrow3-4x+24+6x-x-27-3x=0\)
\(\Leftrightarrow-2x=0\)\(\Leftrightarrow x=0\)
C) \(5-\left(6-x\right)=4\left(3-2x\right)\)\(\Leftrightarrow5-6x+x-12+8x=0\)\(\Leftrightarrow9x-13=0\)\(\Leftrightarrow x=\frac{13}{9}\)
D) \(4\left(x+3\right)=-7x+17\) \(\Leftrightarrow4x+12+7x-17=0\)\(\Leftrightarrow11x-5=0\)\(\Leftrightarrow x=\frac{5}{11}\)
E) \(11x+42-2x=100-9x-22\)\(\Leftrightarrow11x+42-2x-100+9x+22=0\)
\(\Leftrightarrow18x-36=0\)\(\Leftrightarrow x=2\)
F) \(5x+\frac{3}{12}=1+\frac{2x}{9}\)\(\Leftrightarrow180x+9-36-8x=0\)\(\Leftrightarrow172x-27=0\)\(\Leftrightarrow x=\frac{27}{172}\)
TK nka !!! Mk giải tiếp !!
Giúp mình với
a)3y-2y=2y-3
b)3-4x+24+6x=x+27+3x
c)5-(6-x)=4(3-2x)
d)4(x+3)=-7x+17
e)11x+42-2x=100-9x-22
g)2x+3/3 =5-4x/2
f)5x+3/12 = 1+2x/9
h)7x-1/6 = 16-x/5
i)x-3/5 = 6-1-2x/3
k)3x-2/6 - 5=3-2(x+7)/4
3x-7/2 + x+1/3=-16
x-x+1/3 = 2x+1/5
2x-1/3 - 5x+2/7
tìm x
a, \(x^2-4x+\dfrac{1}{4}\)
b, \(8x^2-25\)
c, \(x^2+7x=8\)
d, \(x^3-3x=-27+9x\)
e, x(x-3)-7x=-21
f, \(x^3-2x^2+x-2=0\)
g, \(x^2-4x=-4\)
h, \(x^3-x^2+x=-1\)
\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)
\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)