Tìm x dạng cùng lũy thừa bậc chẵn (2 trường hợp)
x2=3 x2=36 x2=25 2x2+(-20)=55 2.(x-1)2+50=9 -(x+1)2-5=2.(-3).5
Tìm x dạng cùng lũy thừa bậc chẵn (2 trường hợp)
x2=3 x2=36 x2=25 2x2+(-20)=55 2.(x-1)2+50=9 -(x+1)2-5=2.(-3).5
Nhờ các tiền bối giải nhanh giúp mình ,mình đang cần gấp !!!!
X2=3 x2=25
=> X=\(\pm\sqrt{3}\) => x=5
X2=36
=> x=6
2.(x-1)2+50= 9
2.(x-1)2+1= 9
2.(x-1)2= 8
(x-1)2 = 8/2
(x-1)2 = 4
(x-1)2 = (2)2
x-1=(\(\pm\)2)
TH1: x-1= 2 TH2: x-1=-2
x=2+1 x =(-2)+1
x= 3 x = -1
Vậy x\(\in\)\(\left\{3;1\right\}\)
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
Giải phương trình chứa ẩn ở mẫu:
a. (x+1)/(x-2) - (x-1)(x+2) = 2(x2 + 2)/(x2 - 4)
b. (2x+1)/(x-1) = 5(x-1)/(x+1)
c. (x-1)/(x+2) - (x)/(x-2) = (5x-2)/(4 - x2)
d. (x-2)/(2+x)-(3)/(x-2)= 2(x-11)/(x2 - 2)
e. (x-1)/(x+1)-(x2 + x - 2)/(x+1)= (x+1)/(x-1) - x - 2
f. (x+1)/(x-1)-(x-1)/(x+1)=(4)/(x2 - 1)
g. (3)/4(x-5) + (15)/(50-2x2)= - (7)/6(x+5)
h. (12)/(8+x3)= 1 + (1)/(x+2)
k. (x+25)/(2x2 - 50)-(x+5)(x2 - 5x)= (5-x)(2x2 + 10x)
\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x^2+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+2x+x+2-\left(x^2-2x-x+2\right)=2x^2+4\)
\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)
\(\Leftrightarrow6x=2x^2+4\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)
\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)
\(\Leftrightarrow3x^2-13x+4=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)
\(c,\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^2}\)
\(\Leftrightarrow\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{2-5x}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2-5x}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2-2x-x+2-x^2-2x=2-5x\)
\(\Leftrightarrow-5x+2=2-5x\)
\(\Leftrightarrow-5x+5x=2-2\)
\(\Leftrightarrow0=0\)
=>pt luôn có nghiệm với mọi x.
Giải phương trình chứa ẩn ở mẫu:
a. (x+1)/(x-2) - (x-1)(x+2) = 2(x2 + 2)/(x2 - 4)
b. (2x+1)/(x-1) = 5(x-1)/(x+1)
c. (x-1)/(x+2) - (x)/(x-2) = (5x-2)/(4 - x2)
d. (x-2)/(2+x)-(3)/(x-2)= 2(x-11)/(x2 - 2)
e. (x-1)/(x+1)-(x2 + x - 2)/(x+1)= (x+1)/(x-1) - x - 2
f. (x+1)/(x-1)-(x-1)/(x+1)=(4)/(x2 - 1)
g. (3)/4(x-5) + (15)/(50-2x2)= - (7)/6(x+5)
h. (12)/(8+x3)= 1 + (1)/(x+2)
k. (x+25)/(2x2 - 50)-(x+5)(x2 - 5x)= (5-x)(2x2 + 10x)
Tìm x dạng có lũy thừa
1) x^2 - 3 = 22
2) 2x^3 + 5 = 11
3) ( x + 2 ) = 81
4) ( 2x + 1 ) ^2 = 25
5) 5^x + 2 = 625
6) ( 2x - 3 ) ^2 = 36
7) ( 2x - 1 ) ^3 = - 8
( x - 1 ) ^ x +2 = ( x - 1 ) ^ x + 6 với x thuộc Z 19 ^x2 + x = 0
GIÚP MÌNH NHA MÌNH ĐANG CẦN GẤP CẢM ƠN TRUÓC LUÔN NHÁ!
Thực hiện phép tính:
a,(2x- 4)(x+9)
b,(x2 + 4x +3)(x-2)
c,(x-8)(x+8)
d, x2(7x-5)-7(x3- 4x+6)
e,(x2+2)(x2+x+1)
f,(x2+2)(x4-2x2+4)
g,(x-g)(x+9)
h,(x-2)(2x3-x2+1)+(x2+1)+(x2-2x2)(1-2)x
Dễ
Thế
Mà
Cũnhoir
Dc
Ạ
Chịu
Chắc
Phải
Ngu
Lamqs
Mới
Hỏi
Câu
Này
a) x2(x - 5) + 5 - x = 0; b) 3x4 - 9x3 = -9x2 + 27x;
c) x2(x + 8) + x2 = -8x; d) (x + 3)(x2 -3x + 5) = x2 + 3x.
e) 3x(x - 1) + x - 1 = 0;
f) (x - 2)(x2 + 2x + 7) + 2(x2 - 4) - 5(x - 2) = 0;
g) (2x - 1)2 - 25 = 0;
h) x3 + 27 + (x + 3)(x - 9) = 0.
i)8x3 - 50x = 0; k) 2(x + 3)-x2 - 3x = 0;
m)6x2 - 15x - (2x - 5)(2x + 5) =
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Bài 5. Tìm x , biết rằng: a) x(x + 5)(x – 5) – (x + 2)(x2 – 2x + 4) = 3
b) (x – 3)3 – (x – 3)(x2 + 3x + 9) + 9(x + 1)2 = 15
c) (x+5)(x2 –5x +25) – (x – 7) = x3
d) (x+2)(x2 – 2x + 4) – x(x2 + 2) = 4
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
Bài 1 Rút gọn biểu thức
a, [(3x - 2)(x + 1) - (2x + 5)(x2 - 1)] : (x + 1)
b, (2x + 1)2 - 2(2x + 1)(3 - x) + (3 - x)2
c, (x - 1)2 - (x + 1) (x2 - x + 1) - (3x + 1)(1 - 3x)
d, (x2 + 1)(x - 3) - (x - 3)(x2 + 3x + 9)
e, (3x +2)2 + (3x - 2)2 - 2(3x + 2)(3x - 2) + x
Bài 2 Phân tích các đa thức sau thành nhân tử
1, 3(x + 4) - x2 - 4x
2, x2 - xy + x - y
3, 4x2 -25 + (2x + 7)(5 - 2x)
4, x2 + 4x - y2 + 4
5, x3 - x2 - x + 1
6, x3 + x2y - 4x - 4y
7, x3 - 3x2 + 1 - 3x
8, 2x2 + 3x - 5
9, x2 - 7xy + 10y2
10, x3 - 2x2 + x - xy2