a x2 +2x√x+1=8x-1
b x+√9-x2 -x√9-x2=3
c x2+x+12√x+1=36
d x+√17-x2 + x√17-x2=9
Những câu hỏi liên quan
= 26/24 x 27/5 x2x34/9 x 2x2/17
= 13/12 x 27/5 x2 x34/9 x2 x 2/17
= 13/12 x2x2 x27/5 x 34/9 x2/17
= 13/3 x27/5 x 4/9
= 117/5x4/9
= 52/5
Tìm x nguyên để các biểu thức sau nguyên
a) 4x+ 2
______
5x + 1
b) x2 + 3x +9
_________
x + 3
c) x2 + 9
_____
x + 2
a: Để A nguyên thì 4x+2 chia hết cho 5x+1
=>20x+10 chia hết cho 5x+1
=>20x+4+6 chia hết cho 5x+1
=>5x+1 thuộc {1;-1;2;-2;3;-3;6;-6}
=>x thuộc {0;-2/5;1/5;-3/5;2/5;-4/5;1;-7/5}
b: B nguyên
=>x^2+3x+9 chia hết cho x+3
=>9 chia hết cho x+3
=>x+3 thuộc {1;-1;3;-3;9;-9}
=>x thuộc {-2;-4;0;-6;6;-12}
c: Để C nguyên thì x^2+9 chia hết cho x+2
=>x^2-4+13 chia hết cho x+2
=>x+2 thuộc {1;-1;13;-13}
=>x thuộc {-1;-3;11;-15}
Đúng 0
Bình luận (0)
Tìm X
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47) f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100 h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6 k) 17/ (x + 3)3 : 3 – 1 = – 10
e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
Đúng 0
Bình luận (0)
Tìm X
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47) f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100 h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6 k) 17/ (x + 3)3 : 3 – 1 = – 10
Tìm X
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47) f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100 h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6 k) 17/ (x + 3)3 : 3 – 1 = – 10
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)
Đúng 0
Bình luận (0)
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
Đúng 0
Bình luận (0)
1/Rút gọn các biểu thức: a)(x+1)2-(x-1)2-3(x+1)(x-1)
b)5(x+2)(x-2)-(2x-3)2-x2+17
c)(x-3)3-(x-3)(x2+3x+9)+6(x+1)2
2/ Tìm x a) (x+4)2-(x+1)(x-1)=16
b) (x+2)(x2-2x+4)-x(x2+2)=15
Tìm x, biết:a)
x
2
(x - 5) + 5 - x 0; b) 3
x
4
- 9
x
3
-9
x
2
+ 27x;c)
x
2
(x + 8) +
x
2
-8x; d) (x + 3)(
x
2
-3x + 5)
x
2...
Đọc tiếp
Tìm x, biết:
a) x 2 (x - 5) + 5 - x = 0; b) 3 x 4 - 9 x 3 = -9 x 2 + 27x;
c) x 2 (x + 8) + x 2 = -8x; d) (x + 3)( x 2 -3x + 5) = x 2 + 3x.
a. x+1/x-2 - x/x+2 + 8/x2 -4
b. x-3/x+1 - x+2/x-1 + 8x/x2 -1
c. x+2/x2-2x + 2/x2+2x + 3x+2/x2-4
d. 4/x - 12/x2+3x + 5/x+3
a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
c: \(=\dfrac{x+2}{x\left(x-2\right)}+\dfrac{2}{x\left(x+2\right)}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x^2+2x+2x-4+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+7x-2}{x\left(x-2\right)\left(x+2\right)}\)
Đúng 0
Bình luận (0)
a,
\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b,
\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)
Đúng 0
Bình luận (0)