(x+4)\(^2\)-(x+1).(x+1)=16
tìm x
Cho đa thức g (x) = a.x2 - 4.x + 16
Tìm a để x= -2 là nghiệm của đa thức g (x) ?
G(-2)=0
=>4a+8+16=0
=>4a=-24
hay a=-6
đặt g(-2) = 0
=> a.(-2)2 - 4.(-2) + 16 = 0
<=> 4a + 8 + 16 = 0
<=> 4a = -24
<=> a = -6
vậy để x = -2 là nghiệm của g(x) thì a = -6
a)-1/3x+24
b)2x/x^2-16
Tìm các giá trị của x để các phân thức sau có nghĩa:trình bày ra luôn
3x=2y=7z và x- z= 16
tìm x,y,z
\(3x=2y=7z\)
⇒\(\dfrac{x}{14}=\dfrac{y}{21}=\dfrac{z}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{14}=\dfrac{z}{6}=\dfrac{x-z}{14-6}=\dfrac{16}{8}=2\)
⇒\(\left\{{}\begin{matrix}x=14.2=28\\y=21.2=42\\z=6.2=12\end{matrix}\right.\)
B1:tìm x biết a, (-2+x^2)(x^2-2)(x^2-2)(x^2-2)(x^2-2)=1 b, (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4) c,(8x-3)(3x+2)-(4x+7)(x+4)=(4x+1)(5x-1) d, 2x^2+3(x-1)(x+1)=5x(x+1) e, (8-5x)(x+2)+4(x-2)(x+1)=(2+x)(2-x) f, 4(x-1)(x+5)-(x+2)(x+5)=3(x-1)(x+2)
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
Chúng ta sẽ giải từng phương trình một:
a. Đặt , ta có:
1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
Bài 1:
a, (-2+x\(^2\))(x\(^2\)-2)(x\(^2\)-2)(x\(^2\)-2)(x\(^2\)-2)=1
b, (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
c, (8x-3)(3x+2)-(4x+7)(x+4)=(4x+1)(5x-1)
d,2x\(^2\)+3(x-1)(x+1)=5x(x+1)
e,(8-5x)(x+2)+4(x-2)(x+1)=(2+x)(2-x)
f, 4(x-1)(x+5)-(x+2)(x+5)=3(x-1)(x+2)
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1
=>-16x-34=x-1
=>-17x=33
=>x=-33/17
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
e: =>8x+16-5x^2-10x+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6
=>4x^2+16x-20-4x^2-10x+4=0
=>6x=16
=>x=8/3
1:
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x^2+5x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)
\(\Leftrightarrow x^2+5x=0\)
=>x=0 hoặc x=-5
3: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
1 x 2 = ?
1 x 2 x 3 = ?
1 x 2 x 3 x 4 = ?
1 x 2 x 3 x 4 x 5 = ?
1 x 2 x 3 x 4 x 5 x 6 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = ?
1x2= 2 1x2x3=6 1x2x3x4=24 1x2x3x4x5=120 1x2x3x4x5x6=720 1x2x3x4x5x6x7=5040
1x2x3x4x5x6x7x8=40320 1x2x3x4x5x6x7x8x9=362880 1x2x3x4x5x6x7x8x9x10=3628800
1 x 2 = 2
1 x 2 x 3 = 6
1 x 2 x 3 x 4 = 24
1 x 2 x 3 x 4 x 5 = 120
1 x 2 x 3 x 4 x 5 x 6 = 720
1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 362880
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3628800
1 . 2 = 2
1 .2 .3 = 6
1 .2 .3 .4 = 24
1 .2 .3 .4 .5 = 120
1 .2 .3 .4 .5 .6 = 720
1 .2 .3 .4 .5 .6 .7 = 5040
1 .2 .2 .4 .5 .6 .7 .8 = 40320
1 .2 .3 .4 .5 .6 .7 .8 .9 =362880
1 .2 .3 .4 .5 .6 .7 .8 .9 .10 = 3628800
hok tốt
a,x+5/x-1+8/x^2-4x+3=x+1/x-3 b,x-4/x-1-x^2+3/1-x^2+5/x+1=0 c,3x/4-5=3-x/2+5x-1/6 d,(x-2)(x+2)-(x-3)(x+4)-2x+3=0 e,(x-1)^2+2(x+1)=5x+5 g,(x-3)(x+4)x=0
a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)
=>(x+5)(x-3)+8=x^2-1
=>x^2+2x-15+8=x^2-1
=>2x-7=-1
=>x=3(loại)
b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)
=>(x-4)(x+1)+x^2+3+5(x-1)=0
=>x^2-3x-4+x^2+3+5x-5=0
=>2x^2+2x-6=0
=>x^2+x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
e: =>x^2-2x+1+2x+2=5x+5
=>x^2+3=5x+5
=>x^2-5x-2=0
=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)
g: (x-3)(x+4)*x=0
=>x=0 hoặc x-3=0 hoặc x+4=0
=>x=0;x=3;x=-4