p(x),q(x) là số vô tỉ hay hữu tỉ:
P(x)=\(x^6+2x^5-2x^4+x^3+5x^2-6x-1\)
Q(x)=\(x^7-2x^6+2x^5-x^4-5x^3-6x^2-6x+1\)
Những câu hỏi liên quan
P(x) , Q(x) là số vô tỉ hay số hữu tỉ biết:
P(x)=x6 +2x5-2x4+x3+5x2-6x-1
Q(x)=x7 -2x6 +2x5 -x4-5x3- 6x2 -6x+1
:)) đk của x nữa bạn. như vậy thì làm kiểu gì được
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a) x+1/2x-6-4/2x-6 b) 3x-4/6x+3-x-5/6x+3
c) x-1/x-3-3x-8/3-x+3-2x/x-3 d) 3/x+5-5/x-7
e) 3/x+5-5/x-7 f) 2/x-2+3/x+2+5x-18/x2-4
Tìm x:
1) -3.(1-2x) - 4.(1+3x) = -5x + 5
2) 3.(2x - 5) - 6.(1 - 4x) = -3x + 7
3) (1 - 3x) - 2.(3x - 6) = -4x - 5
4) x.(4x - 3) - 2x.(2x - 1) = 5x - 7
5) 3x.(2x - 1) - 6x.(x + 2) = -3x + 4
6) (1 - 2x).3 - 4.(6x - 1) = 7x - 5
7) 6x - 3.(1 - 4x) - 5.(x + 1) = 2x + 7
8) 6.(1 - 3x) - 3.(2x + 5) = -10x + 7
9) 3x.(1 - 2x) + 6x^2 - 7x = 8.(1 - 2x) - 9
10) 2x.(1 + 3x) - 3x.(4 + 2x) = 3x - 4
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
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1) (x+6)(3x-1)+x+6=0
2) (x+4)(5x+9)-x-4=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
4)2x (2x-3)=(3-2x)(2-5x)
5)(2x-7)^2-6(2x-7)(x-3)=0
6)(x-2)(x+1)=x^2-4
7) x^2-5x+6=0
8)2x^3+6x^2=x^2+3x
9)(2x+5)^2=(x+2)^2
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
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8 tháng 9 2021 lúc 14:22
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
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a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
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1. Các hằng đẳng thức sau là đúnga. x^2+6x+9/x^2+3x+3/x+1b. x^2-4/5x^2+13x+6x+2/5x+3c. x^2+5x+4/2x^2+x-3x^2+3x+4/2x^2-5x+3d. x^2-8x+16/16-x^24-x/4+x2. P là đa thức nào để x^2+2x+1/Px^2-1/4x^2-7x+3a. P4x^2+5x-2b. P4x^2+x-3c. P4x^2-x+3d. P4x^2+x+33. Đa thức Q trong đẳng thức 5(y-x)^2/5x^2-5xyx-y/Qa. x+yb. 5(x+y)c. 5(x-y)d. x4. Đa thức Q trong hằng đẳng x-2/2x^2+32x^2-4x/Q là:a. 4x^2+16b. 6x^2-4xc. 4x^3+6xd. khác5. Phân thức 2x+1/2x-3 bằng phân thức:a. 2x^2+x/2x-3b. 2x^2+x/2x^2-3xc. 2x+1/6x-9d. Khá...
Đọc tiếp
1. Các hằng đẳng thức sau là đúng
a. x^2+6x+9/x^2+3=x+3/x+1
b. x^2-4/5x^2+13x+6=x+2/5x+3
c. x^2+5x+4/2x^2+x-3=x^2+3x+4/2x^2-5x+3
d. x^2-8x+16/16-x^2=4-x/4+x
2. P là đa thức nào để x^2+2x+1/P=x^2-1/4x^2-7x+3
a. P=4x^2+5x-2
b. P=4x^2+x-3
c. P=4x^2-x+3
d. P=4x^2+x+3
3. Đa thức Q trong đẳng thức 5(y-x)^2/5x^2-5xy=x-y/Q
a. x+y
b. 5(x+y)
c. 5(x-y)
d. x
4. Đa thức Q trong hằng đẳng x-2/2x^2+3=2x^2-4x/Q là:
a. 4x^2+16
b. 6x^2-4x
c. 4x^3+6x
d. khác
5. Phân thức 2x+1/2x-3 bằng phân thức:
a. 2x^2+x/2x-3
b. 2x^2+x/2x^2-3x
c. 2x+1/6x-9
d. Khác
Câu 5:B
Câu 4: C
Câu 3: D
Câu 2: A
Câu 1: A
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giải các phương trình
1) sqrt{4x-20} +3sqrt{dfrac{x-5}{9}} -dfrac{1}{3}sqrt{9x-45}6
2)sqrt{x+1}+sqrt{x+6}5
3) x^2-6x+sqrt{x^2-6x+7}5
4)sqrt{x+3-4sqrt{x-1}}+sqrt{x+8-6sqrt{x-1}}4
5)sqrt{x^2-dfrac{1}{4}+sqrt{x^2+x+dfrac{1}{4}}}dfrac{1}{2}left(2x^3+x^2+2x+1right)
6)sqrt{3x^2+6x+12}+sqrt{5x^4-10x^2+30}8
7)sqrt{3x^2+6x+7}+sqrt{5x^2+10x+14}4-2x-x^2
Đọc tiếp
giải các phương trình
1) \(\sqrt{4x-20}\) +3\(\sqrt{\dfrac{x-5}{9}}\) \(-\dfrac{1}{3}\sqrt{9x-45}=6\)
2)\(\sqrt{x+1}+\sqrt{x+6}=5\)
3) \(x^2-6x+\sqrt{x^2-6x+7}=5\)
4)\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=4\)
5)\(\sqrt{x^2-\dfrac{1}{4}+\sqrt{x^2+x+\dfrac{1}{4}}}=\dfrac{1}{2}\left(2x^3+x^2+2x+1\right)\)
6)\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
7)\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
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2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
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3)
ĐK: \(x^2-6x+7\geq 0\)
Đặt \(\sqrt{x^2-6x+7}=a(a\geq 0)\) \(\Rightarrow x^2-6x=a^2-7\)
PT trở thành: \(a^2-7+a=5\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow (a-3)(a+4)=0\Rightarrow a=3\) (do \(a\geq 0)\)
\(\Rightarrow \sqrt{x^2-6x+7}=3\)
\(\Rightarrow x^2-6x+7=9\)
\(\Leftrightarrow x^2-6x-2=0\) \(\Rightarrow x=3\pm \sqrt{11}\) (đều thỏa mãn)
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Xem thêm câu trả lời
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0
c. x^2-5x+6=0
<=> x^2-5x=-6
<=> -4x=-6
<=> x=-6/-4
vậy tập nghiệm của pt là s={-6/-4}
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Help me
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0