a,(x+3)(x^2-3x+9)-(x-3)(x^2+3x+9)
b,(x-5)(x^2+5x+25)-(x+5)(x^2-5x+25)
c,(x-4)(x^2+4x+16)-(x+4)(x^2-4x+18)
d,(x-2)(x^2+7x+49)-(x-7)(x^2-7x+49)
A.(X+16/49) +(x+18/47)=(x+20/45)-1
B.(x-3/x-2)+(x-2/x-4)=-1
C.(4x-10).(x+6)=0
D.7x-1/2=5+(9-5x/6)
E.3.(7x-1)=30+9-5
F.(3x-9/x+1)-2=4x/x+1
Giải pt
a)1/x-1-3x²/x³-1=2x/x²+x+1
b) 4x+5/x-1+2x-1/x+1=6
c) x+2/x-3+x-2/x+3=2(x²+6) /x²-9
d) 1/x-2+3=3x-2/x+2
e) 148-x/25+169-x/23+186-x/21+199-x/19=10
f) x+16/49+x+18/47=x+20/45-1
g) (x+2) (x+4) (x+6) (x+8) =-16
h) x³+3x²+3x+2=0
i) 4x/x²-5x+6+3x/x²-7x+6=0
\(a,\)( sửa lại xíu đề cho đúng nhé )
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=-\frac{2x}{x^2+x+1}\)
\(\Rightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1-3x^2=-2x^2+2x\)
\(\Rightarrow x=1\)
\(g,\)\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)=-16\)
\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)=-16\)
Đặt \(x^2+10x+16=a\)
\(\Rightarrow a\left(a+8\right)=-16\)
\(\Rightarrow a^2+8a+16=0\)
\(\Rightarrow\left(a+4\right)^2=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)
\(\Rightarrow x^2+10x+25-25=0\)
\(\Rightarrow\left(x+5\right)^2-\left(\sqrt{5}\right)^2=0\)
\(\Rightarrow\left(x+5-\sqrt{5}\right)\left(x+5+\sqrt{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-5+\sqrt{5}\\x=-5-\sqrt{5}\end{cases}}\)
\(h,\)\(x^3+3x^2+3x+2=0\)
\(\Rightarrow x^3+2x^2+x^2+2x+x+2=0\)
\(\Rightarrow x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x^2+x+1\right)=0\)
Vì \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow x+2=0\Leftrightarrow x=-2\)
1) (4-3x) (10x-5)=0
2) (7-2x) (4+8x) = 0
3) (9-7x) (11-3x) = 0
4) (7-14x) (x-2) = 0
5) (2x+1) (x-3) = 0
6) (8-3x) (-3x+5) = 0
7) (16-8x) (2-6x) = 0
8) (x+4) (6x-12) = 0
9) (11-33x) (x+11) = 0
10) (x-1/4) (x+5/6) = 0
11) (7/8-2x) (3x+1/3) = 0
12) 3x - 2x^2 = 0
13) 5x + 10x^2 = 0
14) 4x + 3x^2 = 0
15) -8x^2 + x =0
16) 10x^2 - 15x = 0
17) x^2 -4 =0
18) 9 - x^2 = 0
19) x^2 -1 = 0
20) (x-3) (2x-1) = (2x-1) ( 2x+3)
21) (5+4x) (-x+2) = (5+4x) (7+5x)
22) (4+x) (x-5) = (3x-8) (x-5) = 0
23) (3x-8) (7-21x) - (9+2x) (7-21x)
24) (10+ 7x) (x+1) = (9x-2)(x-1)
25) (9x-4) (x-1/2) - (x-1/2) (6+x) = 0
26) 9x^2 - 1 = (3x-1) (x+4)
27) (x+7) (3x+1) = 49-x^2
28) (2x+1)^2 = (x-1)^2
29)x^3- 5x^2+6x = 0
30) 3x^2 + 5x + 2 = 0
Giảii giúpp mìnhh đyy mọii ngườii .
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
Thu gọn biểu thức a)x^2-4x(3x-4)+7x-5 b)7x(x^2-5)-3x^2y(xy-6y^2) c)(5x+4)(2x-7) d)4x(2x-3)-5x(x-2)+x^2(3-x)
a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)
\(=x^2-12x^2+16x+7x-5\)
\(=-11x^2+23x-5\)
b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)
\(=7x^3-35x-3x^3y^2+18x^2y^3\)
c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)
\(=10x^2-35x+8x-28\)
\(=10x^2-27x-28\)
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`
Bài 1: Tìm x biết :
a) (x+2)^2 - (3x-7)^2=0
b) (4x+1) -(5x-3)^2=0
c) 25(x-3)^2 - 49(2x+1)^2=0
d) 9(3x-2)^2=121(1-4x)^2
e) (x-5/4)^2=(5x+1/2)^2
a ) 5x ( 2 - 3x ) = 4 - 6x
b ) ( 5 - x ) ( 2 + 3x ) = 4 - 9x2
c ) 25 – x2 = 4x ( 5 + x )
d ) x2 - 2x + 1 = 3x ( x - 1 )
e ) 4 ( 7x - 3 ) = 7x2 - 3x
g ) ( x - 5 )4 + ( x - 3 )4 = 16 .
Phân tích các đa thức sau thành nhân tử.
1) a^2+ab+2b-4 2) x^3-x 3) x^2-6x+8 4) ab+b^2-3a-3b 5) x^3-4x^2-8x+8
6)9x^2+6x-8 7)x^2-y^2-4x+4 8)5x^3-10x^2+5x 9) 3x^2-8x+4 10) 4x^2-4x-3
11) x^2-7x+12 12)x^2-5x-14 13) 3x^2-7x+2 14) a.(x^2+1)-x.(a^2-1) 15) x^4+4
16) (x+2).(x+3).(x+4).(x+5)-24 17) (a+1).(a+3).(a+5).(a+7)+15
a. \(\frac{x-3}{x-2}+\frac{x-2}{x-4}=-1\)
b.\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
c.\(\frac{7x-1}{2}=5+\frac{9-5x}{6}\)
d.\(\frac{3x-9}{x+1}-2=\frac{4x}{x+1}\)
c) \(\dfrac{7x-1}{2}=5+\dfrac{9-5x}{6}\)
\(\Leftrightarrow\dfrac{6\left(7x-1\right)}{12}=\dfrac{5\cdot12}{12}+\dfrac{2\left(9-5x\right)}{12}\)
\(\Rightarrow42x-6=60+18-10x\)
\(\Leftrightarrow52x-84=0\)
\(\Leftrightarrow x=\dfrac{21}{13}\)
Vậy....
d) tương tự
a) \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)ĐKXĐ : \(x\ne2;4\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Leftrightarrow\dfrac{2x^2-11x+16}{x^2-6x+8}=-1\)
\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)( thỏa mãn ĐKXĐ )
Vậy....
b) \(\dfrac{x+16}{49}+\dfrac{x+18}{47}=\dfrac{x+20}{45}-1\)
\(\Leftrightarrow\dfrac{x+16}{49}+1+\dfrac{x+18}{47}+1=\dfrac{x+20}{45}-1+1+1\)
\(\Leftrightarrow\dfrac{x+16+49}{49}+\dfrac{x+18+47}{47}=\dfrac{x+20+45}{45}\)
\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}=\dfrac{x+65}{45}\)
\(\Leftrightarrow\left(x+65\right)\left(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\right)=0\)
Vì \(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\ne0\)
\(\Leftrightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
Vậy....