giải các phương trình sau
b) ( x + 2)(x-3)(x + 1)(x + 6) = - 36
c) 6x4 - 35x3 + 62x2 - 35x + 6 = 0
d) x4 + x3 -4x2 + x + 1 =0
e) x4 + ( x-1)4 = 97
f) x4 -5x3 + 10x2 -10x + 4 =0
g)(x+3)4 + (x+5)4=16
giúp e với ạ >< cần gấp ạ ><
Những câu hỏi liên quan
giải phương trình sau:
a. (9x2-4)(x+1) = (3x+2) (x2-1)
b. (x-1)2-1+x2 = (1-x)(x+3)
c. (x2-1)(x+2)(x-3) = (x-1)(x2-4)(x+5)
d. x4+x3+x+1=0
e. x3-7x+6 = 0
f. x4-4x3+12x-9 = 0
g. x5-5x3+4x = 0
h. x4-4x3+3x2+4x-4 = 0
m.n jup vs
Giải các phương trình sau:a, (9x2 - 4)(x + 1) (3x +2)(x2 - 1)b, (x - 1)2 - 1 + x2 (1 - x)(x + 3)c, (x2 - 1)(x + 2)(x - 3) (x - 1)(x2 - 4)(x + 5)d, x4 + x3 + x + 1 0e, x3 - 7x + 6 0f, x4 - 4x3 + 12x - 9 0g, x5- 5x3 + 4x 0h, x4 - 4x3 + 3x2 + 4x - 4 0
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Giải các phương trình sau:
a, (9x2 - 4)(x + 1) = (3x +2)(x2 - 1)
b, (x - 1)2 - 1 + x2 = (1 - x)(x + 3)
c, (x2 - 1)(x + 2)(x - 3) = (x - 1)(x2 - 4)(x + 5)
d, x4 + x3 + x + 1 = 0
e, x3 - 7x + 6 = 0
f, x4 - 4x3 + 12x - 9 = 0
g, x5- 5x3 + 4x = 0
h, x4 - 4x3 + 3x2 + 4x - 4 = 0
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
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b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
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Bài 1: Giải các phương trình dưới đây1) x2 - 9 (x - 3)(5x +2)2) x3 - 1 (x - 1)(x2 - 2x +16)3) 4x2 (x - 1) - x + 1 04) x3 + 4x2 - 9x - 36 05) (3x + 5)2 (x - 1)26) 9 (2x + 1)2 4 (x - 5)27) x2 + 2x 158) x4 + 5x3 + 4x2 09) (x2 - 4) - (x - 2)(3 - 2x) 010) (3x + 2)(x2 - 1) (9x2 - 4) (x + 1)11) (3x - 1)(x2 + 2) (3x - 1)(7x - 10)12) (2x2 + 1) (4x - 3) (x - 12)(2x2 + 1)
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Bài 1: Giải các phương trình dưới đây
1) x2 - 9 = (x - 3)(5x +2)
2) x3 - 1 = (x - 1)(x2 - 2x +16)
3) 4x2 (x - 1) - x + 1 = 0
4) x3 + 4x2 - 9x - 36 = 0
5) (3x + 5)2 = (x - 1)2
6) 9 (2x + 1)2 = 4 (x - 5)2
7) x2 + 2x = 15
8) x4 + 5x3 + 4x2 = 0
9) (x2 - 4) - (x - 2)(3 - 2x) = 0
10) (3x + 2)(x2 - 1) = (9x2 - 4) (x + 1)
11) (3x - 1)(x2 + 2) = (3x - 1)(7x - 10)
12) (2x2 + 1) (4x - 3) = (x - 12)(2x2 + 1)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
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1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
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7.
\(\Leftrightarrow x^2+2x-15=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)
\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0;x=-4\end{matrix}\right.\)
9.\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(3-2x\right)\)
\(\Leftrightarrow x+2=3-2x\)
\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)
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Giải các phương trình sau:a)
x
+
2
3
+
x
+
1
3
0
;
b)
2
x
4
+
3
x
2
−
5
0
;
c)
x...
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Giải các phương trình sau:
a) x + 2 3 + x + 1 3 = 0 ;
b) 2 x 4 + 3 x 2 − 5 = 0 ;
c) x 4 − 8 x 3 − 9 x 2 = 0 ;
d) x 3 − 4 x 2 + 4 − x = 0 .
giải các phương trình sau
b) ( x + 2)(x-3)(x + 1)(x + 6) = - 36
c) 6x4 - 35x3 + 62x2 - 35x + 6 = 0
d) x4 + x3 -4x2 + x + 1 =0
e) x4 + ( x-1)4 = 97
f) x4 -5x3 + 10x2 -10x + 4 =0
g)(x+3)4 + (x+5)4=16
b: \(\Leftrightarrow\left(x^2+3x+2\right)\left(x^2+3x-18\right)=-36\)
\(\Leftrightarrow\left(x^2+3x\right)^2-16\left(x^2+3x\right)=0\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x-16\right)=0\)
hay \(x\in\left\{0;-3;\dfrac{-3+\sqrt{73}}{2};\dfrac{-3-\sqrt{73}}{2}\right\}\)
c: \(\Leftrightarrow6x^4-18x^3-17x^3+51x^2+11x^2-33x-2x+6=0\)
\(\Rightarrow\left(x-3\right)\left(6x^3-17x^2+11x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^3-12x^2-5x^2+10x+x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(6x^2-5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(3x-1\right)\left(2x-1\right)=0\)
hay \(x\in\left\{3;2;\dfrac{1}{3};\dfrac{1}{2}\right\}\)
d: \(\Leftrightarrow\left(x-1\right)^2\cdot\left(x^2+3x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{-3+\sqrt{5}}{2};\dfrac{-3-\sqrt{5}}{2}\right\}\)
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Cho $x^2-3x+1=0$x2−3x+1=0
Tính $A=\frac{\left(x^4+x^3-10x^2+x+2015\right)\left(x^4+x^2+1\right)+x^4+3x^2+1}{x^4+x^2+1}$A=(x4+x3−10x2+x+2015)(x4+x2+1)+x4+3x2+1x4+x2+1
Các bn giải giúp mik với, ko pik đề có bị sai ko mà lm ko đc
Bài 1: Giải phương trình:
a) ( x+1)2 (x+2) + ( x – 1)2 ( x- 2) = 12
b) x4 + 3x3 + 4x2 + 3x + 1 = 0
c) x5 – x4 + 3x3 + 3x2 –x + 1 = 0
Bài 2: Chứng minh rằng các phương trình sau vô nghiệm
a) x4 – x3 + 2x2 – x + 1 = 0
b) x4 + x3 + x2 + x + 1 = 0
c) x4 – 2x3 +4x2 – 3x +2 = 0
d) x6+ x5+ x4 + x3 + x2 + x + 1 = 0
1.
a/ \(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)+\left(x-1\right)\left(x^2-3x+2\right)-12=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2\right)+3x\left(x+1\right)-3x\left(x-1\right)+\left(x-1\right)\left(x^2+2\right)-12=0\)
\(\Leftrightarrow2x\left(x^2+2\right)+6x^2-12=0\)
\(\Leftrightarrow x^3+3x^2+2x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+6\right)=0\Rightarrow x=1\)
b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(x^2+\frac{1}{x^2}+3\left(x+\frac{1}{x}\right)+4=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(t^2-2+3t+4=0\Rightarrow t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-1\\x+\frac{1}{x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\left(vn\right)\\x^2+2x+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
1c/
\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4-2x^3+5x^2-2x+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^4-2x^3+x^2+x^2-2x+1+3x^2=0\)
\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+3x^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x-1=0\\x=0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn
Vậy pt có nghiệm duy nhất \(x=-1\)
2.
a. \(x^4-x^3+x^2+x^2-x+1=0\)
\(\Leftrightarrow x^2\left(x^2-x+1\right)+x^2-x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(vn\right)\\x^2-x+1=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
Vậy pt vô nghiệm
b.
\(x^4+x^3+x^2+x+1=0\)
\(\Leftrightarrow x\left(x^3+1\right)+x^3+1+x^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)+x^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2-x+1\right)+x^2=0\)
Mà \(\left\{{}\begin{matrix}\left(x+1\right)^2\left(x^2-x+1\right)\ge0\\x^2\ge0\end{matrix}\right.\)
Nên dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\) ko tồn tại x thỏa mãn
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Giải hộ e bài này với ai 👍
Câu 1 : a, 4x2 -3x-1=0 / d, 4x4-5x2+1=0
b, x2 - (1+căn 5)x + căn 5= 0 / e,x2 +3=|4x| / f, 2x + 5cănx +3 =0 / g, (x2 +x +1 ).(x2+x+2)=2 / h, x4-5x2+4=0
c, x4 + x2 -20=0 / k, x phần x2-1 -- 1 phần 2(x+1) = 1phan 2
Bài 2: Tìm x
a) (x-2)2-(2x+3)20 d) x2.(x+1)-x.(x+1)+x.(x-1)0
b) 9.(2x+1)2-4.(x+1)20 e) (x-2)2-(x-2).(x+2)0
c) x3-6x2+9x0 g) x4-2x2+10
h) 4x2+y2-20x-2y+260 i) x2-2x+5+y2-4y0
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Bài 2: Tìm x
a) (x-2)2-(2x+3)2=0 d) x2.(x+1)-x.(x+1)+x.(x-1)=0
b) 9.(2x+1)2-4.(x+1)2=0 e) (x-2)2-(x-2).(x+2)=0
c) x3-6x2+9x=0 g) x4-2x2+1=0
h) 4x2+y2-20x-2y+26=0 i) x2-2x+5+y2-4y=0