a, 4x : 4 = 64
b, 2x . 4 = 128
c, x 17 = x
d, x . x 2 . x3 = 64
e, 5x + 2 + 5 x + 1 = 150
a) A = -3x(x-5) +3( x2 -4x) -3x-10
b) B = 4x( x2 -7x +2) – 4( x3 -7x2 +2x -5)
c) C = 5x( x2 – x) – x2( 5x-5) -15
d) D = 7( x2 -5x+3)- x( 7x-35) -14
e) E = x2 - 4x - x( x-4) -15
Tìm x, biết:
a) 2(5x-8)-3(4x-5) = 4(3x-4) + 11;
b) 2 x ( 6 x - 2 x 2 ) + 3 x 2 ( x - 4 ) = 8;
c) 2 ( x 3 - 1 ) - 2 x 2 ( x + 2 x 4 ) + ( 4 x 5 + 4 ) x = 6;
d)(2x)2(4x-2)-(x3 -8x2) = 15.
a) x = 2 7 b) x = 2.
c) x = 2 d) x = 1.
Bài 1: Giải các phương trình: a)(5x^ 2 -45).( 4x-1 5 - 2x+1 3 )=0 b) (x^ 2 -2x+6).(2x-3)=4x^ 2 -9 d) 3 5x-1 + 2 3-5x = 4 (1-5x).(5x-3) c) (2x + 19)/(5x ^ 2 - 5) - 17/(x ^ 2 - 1) = 3/(1 - x) e) 3/(2x + 1) = 6/(2x + 3) + 8/(4x ^ 2 + 8x + 3) (x^ 2 -3x+2).(x^ 2 -9x+20)=40 (2x + 5)/95 + (2x + 6)/94 + (2x + 7)/93 = (2x + 93)/7 + (2x + 94)/6 + (2x + 95)/5 Bài 2: Giải các phương trình sau: g) a) (x + 2) ^ 2 + |5 - 2x| = x(x + 5) + 5 - 2x b) (x - 1) ^ 2 + |x + 21| - x ^ 2 - 13 = 0 d) |3x + 2| + |1 - 2x| = 5 - |x| c) |5 - 2x| = |1 - x| Bài 3: Cho biểu thức A = ((x + 2)/(x + 3) - 5/(x ^ 2 + x - 6) + 1/(2 - x)) / ((x ^ 2 - 5x + 4)/(x ^ 2 - 4)) a) Rút gọn A. b) Tim x de A = 3/2 c) Tìm giá trị nguyên c dot u a* d hat e A có giá trị nguyên. B = ((2x)/(2x ^ 2 - 5x + 3) - 5/(2x - 3)) / (3 + 2/(1 - x)) Bài 4: Cho biểu thức a) Rút gọn B. b) Tim* d tilde e B>0 . c) Tim* d hat e B= 1 6-x^ 2 . Bài 5: Cho biểu thức H = (2/(1 + 2x) + (4x ^ 2)/(4x ^ 2 - 1) - 1/(1 - 2x)) / (1/(2x - 1) - 1/(2x + 1)) a) Rút gọn H. b) Tìm giá trị nhỏ nhất của H. c)Tim* d vec e bi vec e u thic H= 3 2
TÌm x, biết:
a, 512-(128-5x)=3x+12
b, (2x-1)+(4x-2)+...+(400x-200)=5+10+...+1000
c, (x+2)+(4x+4)+(7x+6)+...+(25x+18)+28x+20)=1560
d, x+4x+5x+9x+14x+...+97x=500
e, 720-[41-(2x-5)]=23.5
f, 697:\(\dfrac{15x+364}{x}\)=17
a)
\(512-\left(128-5x\right)=3x+12\\ 512-128+5x=3x+12\\ 384+5x=3x+12\\ 5x-3x=12-384\\ 2x=-372\\ x=\left(-372\right):2\\ x=-186\)
b)
\(\left(2x-1\right)+\left(4x-2\right)+...+\left(400x-200\right)=5+10+...+1000\\ \left(2x+4x+...+400x\right)-\left(1+2+...+200\right)=5+10+...+1000\\ x\left(2+4+...+400\right)=\left(5+10+...+1000\right)+\left(1+2+...+200\right)\\ 2x\cdot\left(1+2+...+200\right)=5\cdot\left(1+2+...+200\right)+1\cdot\left(1+2+...+200\right)\\ 2x\cdot\left(1+2+...+200\right)=\left(5+1\right)\cdot\left(1+2+...+200\right)\\ 2x\cdot\left(1+2+...+200\right)=6\cdot\left(1+2+...+200\right)\\ \Rightarrow2x=6\\ x=6:2\\ x=3\)
c)
\(\left(x+2\right)+\left(4x+4\right)+\left(7x+6\right)+...+\left(25x+18\right)+\left(28x+20\right)=1560\\ \left(x+4x+7x+...+25x+28x\right)+\left(2+4+6+...+18+20\right)=1560\\ x\left(1+4+7+...+25+28\right)+110=1560\\ 145x+110=1560\\ 145x=1560-110\\ 145x=1450\\ x=1450:145\\ x=10\)
d)
\(x+4x+5x+9x+14x+...+97x=500\\ x\left(1+4+5+9+14+...+97\right)=500\)
Dãy số trong ngoặc có quy luật: Số thứ \(n\) bằng số thứ \(n-1\) cộng số thứ \(n-2\)
Suy ra dãy số đó là: \(1+4+5+9+14+23+37+60+97=250\)
Thế vào ta được:
\(250x=500\\ x=500:250\\ x=2\)
e)
\(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\\ 720-41+\left(2x-5\right)=8\cdot5\\ 720-41+2x-5=40\\ \left(720-41-5\right)+2x=40\\ 674+2x=40\\ 2x=40-674\\ 2x=-634\\ x=\left(-634\right):2\\ x=-317\)
f)
\(697:\dfrac{15x+364}{x}=17\\ \dfrac{15x+364}{x}=697:17\\ \dfrac{15x+364}{x}=41\\ \dfrac{15x+364}{x}\cdot x=41x\\ 15x+364=41x\\ 364=41x-15x\\ 364=26x\\ x=364:26\\ x=14\)
bài 7
4x3 + 12 = 120
b, ( x - 4 )2 = 64
c, ( x + 1 )3 - 2 = 52
d, 136 - ( x + 5)2 = 100
e, 4x = 16
f, 7x. 3 - 147 = 0
g, 2x+3 - 15 = 17
h, 52x-4. 4 = 102
i, (32 - 4x)(7 - x) = 0
k, ( 8 - x)(10 - 2x) = 0
m, 3x + 3x+1 = 108
n, 5x+2 + 5x+1 = 750
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
1) (\(\dfrac{1}{2}\)x + 3)*(x2- 4x- 6)
2) (6x2 -9x +15)*(\(\dfrac{2}{3}\)x+1)
3) (3x2 -x+5)*(x3+5x-1)
4) (x-1)*(x+1)*(x-2)
5) D=(x-7)*(x+5)-(x-4)*(x+3)
6) E= 4x*(x2-x-1)-(x+3)*(x2-2)
7) F= 5x*(x-3)*(x-1)-4x*(x2-2x)
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
a,4x-10=0 b, 7-3x=9-x c, 2x-(3-5x) = 4(x+3)
d, 5-(6-x)=4(3-2x) e, 4(x+3)=-7x+17 f, 5(x-3) - 4=2(x-1)+7
g, 5(x-3)-4=2(x-1)+7 h,4(3x-2)-3(x-4)=7x+20
`a,4x-10=0 `
`<=> 4x=10`
`<=>x=10/4`
`<=>x=5/2`
`b, 7-3x=9-x `
`<=>-3x+x=9-7`
`<=>-2x=2`
`<=>x=-1`
`c, 2x-(3-5x) = 4(x+3)`
`<=>2x-3+5x=4x+12`
`<=>2x+5x-4x=12+3`
`<=>3x=15`
`<=>x=5`
`d, 5-(6-x)=4(3-2x) `
`<=>5-6+x=12-8x`
`<=>x+8x=12-5+6`
`<=>9x=13`
`<=>x=13/9`
`e, 4(x+3)=-7x+17 `
`<=>4x+12=-7x+17`
`<=>4x+7x=17-12`
`<=>11x=5`
`<=>x=5/11`
`f, 5(x-3) - 4=2(x-1)+7`
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`g, 5(x-3)-4=2(x-1)+7 `
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`h,4(3x-2)-3(x-4)=7x+20`
`<=>12x-8-3x+12=7x+20`
`<=>12x-3x-7x=20+8+12`
`<=>2x=40`
`<=>x=20`
a) x(4x+3y)−(y−2x)2
b) (3+x)(x−3)−(x−1)(x2−3)
c)−2(x−3)2+(x+1)(5x−1)
d) (2x+1)(4x2−2x+1)−3x2(x−2)
e) (3x2+19x+20):(3x+4)
f) (7x2+x3+12x−6):(x2+4x−3)
\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)
Tìm x, biết:
A) (x -1)^3+(2-x)(4+2x+x^2)+3x(x+2)=17
B) (x+2)(x^2-2x+4)-x(x^2-2)=15
C) (5x-2)(5x+2)-(5x+3)(5x-4)=8
D) (4x-3)(4x+2)+(4x+5)(1-4x)=2.5^2
1. Dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng :
a) x2y3/5 = 7x3y4/35xy
b) x3 - 4x/10-5x = -x2-2x/5
c)x + 2/ x-1 = (x+2)(x+1)/ x2-1
d) x2 - x - 2/ x+1 = x2 - 3x +2/ x-1
e) x3+8/ x2-2x+4 = x+2
a: \(\dfrac{7x^3y^4}{35xy}=\dfrac{7xy\cdot x^2y^3}{7xy\cdot5}=\dfrac{x^2y^3}{5}\)
b: \(\dfrac{x^3-4x}{10-5x}=\dfrac{-x\left(x-2\right)\left(x+2\right)}{5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}\)
c: \(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x+2}{x-1}\)
d: \(\left(x^2-x-2\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x^2-3x+2\right)\left(x+1\right)\)
=>\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)
e: \(\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)