Bài 4. Tìm x, biết:
a) (2x + 1)^2 - 4(x + 2)^2 = 9
b) (x + 3)^2 - (x - 4)( x + 8) = 1
c) 3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36
Tìm x, biết:
a) (2x + 1)^2 - 4(x + 2)^2 = 9
b) (x + 3)^2 - (x - 4)( x + 8) = 1
c) 3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36
d)(x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1
a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Rightarrow\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\\ \Rightarrow\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\\ \Rightarrow-3\left(4x+5\right)=9\\ \Rightarrow4x+5=-3\\ \Rightarrow4x=-8\\ \Rightarrow x=-2\)
\(b) (x+3)^2-(x-4)(x+8)=1 <=>x^2+6x=9-(x^2+8x-4x-32)=0 \)
\(.<=> X^2+6x+9-x^2-8x+4x+32-1=0\)
\(<=>2x=-40<=>x=-20\)
=> ptrình có tập nghiêm S={-20}
c) 3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36
\(<=>3(x^2+4x+4)+(4x^2-4x+1)-7(x^2-9)=36\)
\(<=>3x^2+12x+12+4x^2-4x+1-7x^2+49=0\)
\(<=>8x=-62<=>x=7,75\)
=> ptrình có tập nghiệm S={7,75}
d)d)(x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1
\(<=> x^3+3x^2+9x-3x^2-9x-27-x(x^2-4)=1\)
\(<=>x^3+3x^2+9x-3x^2-9x-27-x^3+4x-1=0\)
\(<=> 4x=28<=> x=7\)
=> ptrình có tập nghiệm S={7}
tìm x biết:
a) (x-3)2-4=0
b) x2-2x=24
c) (x+4)2-(x+1)(x-1)=16
d) (2x+1)2-4(x-1)2=9
e) (x+3)2-(x-4)(x+8)=1
f) (2x-1)2+(x+3)2-5(x+7)(x-7)=0
g) 3(x+2)2+(2x-1)2-7(x+3)(x-3)=36
- Gửi lẻ câu hỏi ra nha bạn 2 3 câu 1 lần thôi .
a) (x-3)2-4=0
⇒ (x-3)2=4
⇒ hoặc x-3=2⇒x=5
hoặc x-3=-2⇒x=1
c) (x+4)2-(x+1)(x-1)=16
⇒ x2+8x+16-x2+1=16
⇒ 8x+17=16
⇒ 8x=-1
⇒ x=-1/8
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
Bài 4. Tìm số nguyên x , biết:
a) |x - 2|= 0 b) |x + 3|= 1 c) -3 |4 - x|= -9 d) |2x + 1|= -2
Bài 5. Tìm số nguyên x, biết:
a) (x + 3)mũ 2 = 36 b) (x + 5)mũ 2 =100 c) (2x - 4)mũ 2 = 0 d) (x - 1)mũ 3 = 27
bài 3: tìm x , biết:
a)(x+3).(2x-1)-(x-3).(x+1)=0
b)(x+4).(2x-3)-3.(x-2).(x+2)=0
c)x.(x-5).(x+5)-(x+2).(x2-2x+4)=17
a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
tìm x biết:
a) x - 3= (3 - x)^2
b) x^3 + 3/2x^2 + 3/4x + 1/8 = 1/64
c) 8x^3 - 50x = 0
d) (x - 2) (x^2 + 2x + 7) + 2(x^2 - 4) - 5(x - 2) = 0
e) x(x + 3) - x^2 - 3x = 0
f) x^3 + 27 + (x + 3) (x - 9) = 0
Giúp mik vs ạ
a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
bài 1 : tìm x
a)(2x+1)-4(x+2)2 =99
b)(x+3)2-(x-4)(x+8)=1
c)3(x+2)2+(2x-1)2-7(x+3)(x-3)=36
a) (2x + 1)2 - 4(x + 2)2 = 99
=> 4x2 + 4x + 1 - 4(x2 + 4x + 4) = 99
=> 4x2 + 4x + 1 - 4x2 - 16x - 16 = 99
=> -12x = 114
=> x = -9,5
b) (x - 3)2 - (x - 4)(x + 8) = 1
=> x2 - 6x + 9 - (x2 + 4x - 32) = 1
=> x2 - 6x + 9 - x2 - 4x + 32 = 1
=> -10x = -40
=> x = 4
c) 3(x + 2)2 + (2x - 1)2 - 7(x - 3)(x + 3) = 36
=> 3(x2 + 4x + 4) + 4x2 - 4x + 1 - 7(x2 - 9) = 36
=> 3x2 + 12x + 12 + 4x2 - 4x + 1 - 7x2 + 63 = 36
=> 8x = -40
=> x = -5
a) ( 2x + 1 ) - 4( x + 2 )2 = 99
<=> 4x2 + 4x + 1 - 4( x2 + 4x + 4 ) = 99
<=> 4x2 + 4x + 1 - 4x2 - 16x - 16 = 99
<=> -12x - 15 = 99
<=> -12x = 114
<=> x = -114/12 = -19/2
b) ( x + 3 )2 - ( x - 4 )( x + 8 ) = 1
<=> x2 + 6x + 9 - ( x2 + 4x - 32 ) = 1
<=> x2 + 6x + 9 - x2 - 4x + 32 = 1
<=> 2x + 41 = 1
<=> 2x = -40
<=> x = -20
c) 3( x + 2 )2 + ( 2x - 1 )2 - 7( x + 3 )( x - 3 ) = 36
<=> 3( x2 + 4x + 4 ) + 4x2 - 4x + 1 - 7( x2 - 9 ) = 36
<=> 3x2 + 12x + 12 + 4x2 - 4x + 1 - 7x2 + 63 = 36
<=> 8x + 76 = 36
<=> 8x = -40
<=> x = -5
Bài làm :
Sửa đề : \(a,\left(2x+1\right)^2-4\left(x+2\right)^2=99\)
\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)=99\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=99\)
\(\Leftrightarrow\left(4x^2-4x^2\right)+\left(4x-16x\right)=99+16-1\)
\(\Leftrightarrow-12x=144\)
\(\Leftrightarrow x=-9.5\)
\(b,\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2+6x+9-\left(x^2+8x-4x-32\right)=1\)
\(\Leftrightarrow x^2+6x+9-x^2-8x+4x+32=1\)
\(\Leftrightarrow\left(x^2-x^2\right)+\left(6x-8x+4x\right)=1-32-9\)
\(\Leftrightarrow2x=-40\)
\(\Leftrightarrow x=-20\)
\(c,3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-4x+1-7\left(x^2-9\right)=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\)
\(\Leftrightarrow\left(3x^2+4x^2-7x^2\right)+\left(12x-4x\right)=36-63-1-12\)
\(\Leftrightarrow8x=-40\)
\(\Leftrightarrow x=-5\)
1. Tìm số tự nhiên x biết 1) 3^x . 3=243 2) 7.2^x=56 3) x^3=8 4)x^20=x 5) 2^x-15=17 6) (2x+1)^3=9.81 7) x^6:x^3=115 8) ( 2x-15)^5=(2x-15)^3 9) 3^x+2-5.3^x=36 10) 7.4^x-1+4^x+1=23 Giúp tớ bài này với
1: =>3^x=81
=>x=4
2: =>2^x=8
=>x=3
3: =>x^3=2^3
=>x=2
4: =>x^20-x=0
=>x(x^19-1)=0
=>x=0 hoặc x=1
5: =>2^x=32
=>x=5
6: =>(2x+1)^3=9^3
=>2x+1=9
=>2x=8
=>x=4
7: =>x^3=115
=>\(x=\sqrt[3]{115}\)
8: =>(2x-15)^5-(2x-15)^3=0
=>(2x-15)^3*[(2x-15)^2-1]=0
=>2x-15=0 hoặc (2x-15)^2-1=0
=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1
=>x=15/2 hoặc x=8 hoặc x=7
1. Tìm số tự nhiên x biết:
1) \(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
_____
2) \(7.2^x=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
_____
3) \(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=3\)
_____
4) \(x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x=1\)
5) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
_____
6) \(\left(2x+1\right)^3=9.81\)
\(\left(2x+1\right)^3=729=9^3\)
\(\rightarrow2x+1=9\)
\(2x=9-1\)
\(2x=8\)
\(x=8:2\)
\(\Rightarrow x=4\)
_____
7) \(x^6:x^3=125\)
\(x^3=125\)
\(x^3=5^3\)
\(\Rightarrow x=5\)
_____
8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)
_____
9) \(3^{x+2}-5.3^x=36\)
\(3^x.\left(3^2-5\right)=36\)
\(3^x.\left(9-5\right)=36\)
\(3^x.4=36\)
\(3^x=36:4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
_____
10) \(7.4^{x-1}+4^{x+1}=23\)
\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)
\(4^{x-1}.\left(7+4^2\right)=23\)
\(4^{x-1}.\left(7+16\right)=23\)
\(4^{x-1}.23=23\)
\(4^{x-1}=23:23\)
\(4^{x-1}=1\)
\(4^{x-1}=4^1\)
\(\rightarrow x-1=0\)
\(x=0+1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
Tìm x, biết:
a, (x+8).(x+6)-x^2=104
b, (x+1).(x+2)-(x-3).(x+4)=6
c, 3.(2x-1).(x+2)-2.(3x+2).(x-4)=5
a: \(\Leftrightarrow14x=56\)
hay x=4