\(-\)\(x^2\)\(+\)6x\(-\)4 Tìm...
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\(-\)\(x^2\)\(+\)6x\(-\)4 Tìm nghiệm
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TN
0
TN
0
2 tháng 12 2023
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
S
1
CD
8 tháng 12 2021
a, <=> x2 -2x +1 + 5x -x2 =8
<=> 3x +1 =8
<=> 3x = 7
<=> x= 7/3
b, thiếu đề
c, <=> 2x3 -1 + 2x(4 -x2) = 7
<=> 2x3 + 8x -23 = 8
<=> 8x =8
<=> x=1
W
3
5 tháng 10 2021
Với \(x\ge\dfrac{1}{6}\Leftrightarrow A=5x^2-6x+1-1=5x^2-6x\)
\(A=5\left(x^2-2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{9}{5}=5\left(x-\dfrac{3}{5}\right)^2-\dfrac{9}{5}\ge-\dfrac{9}{5}\\ A_{min}=-\dfrac{9}{5}\Leftrightarrow x=\dfrac{3}{5}\left(1\right)\)
Với \(x< \dfrac{1}{6}\Leftrightarrow A=5x^2+6x-1-1=5x^2+6x-2\)
\(A=5\left(x^2+2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{19}{5}=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{19}{5}\ge-\dfrac{19}{5}\\ A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\)
5 tháng 10 2021
Với \(x\ge\dfrac{1}{3}\Leftrightarrow B=9x^2-6x-4\left(3x-1\right)+6=9x^2-18x+10\)
\(B=9\left(x^2-2x+1\right)+1=9\left(x-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=1\left(1\right)\)
Với \(x< \dfrac{1}{3}\Leftrightarrow B=9x^2-6x+4\left(3x-1\right)+6=9x^2+6x+2\)
\(B=\left(9x^2+6x+1\right)+1=\left(3x+1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=-\dfrac{1}{3}\left(2\right)\)
\(\left(1\right)\left(2\right)\Leftrightarrow B_{min}=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
M
5 tháng 11 2017
1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3)
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27
<=> - 3x^2 - 18x - 22 = 0
<=> 3x^2 + 18x + 22 = 0
5 tháng 11 2017
Nửa chu vi mảnh đất là:
120 : 2 = 60 (m)
Chiều dài hơn chiều rộng là:
5 + 5 = 10 (m)
Chiều rộng là:
( 60 - 10 ) : 2 = 25 (m)
Chiều dài là:
25 + 10 = 35 (m)
Diện tích là:
25 35 = 875 ( )
B
1
NH
21 tháng 6 2023
a)
`4(x-2)^2 =4`
`<=>(x-2)^2 =1`
`<=>x-2=1` hoặc `x-2=-1`
`<=>x=3` hoặc `x=1`
b)
`5(x^2 -6x+9)=5`
`<=>(x-3)^2 =1`
`<=>x-3=1`hoặc `x-3=-1`
`<=>x=4` hoặc `x=2`
c)
`4x^2 +4x+1=0`
`<=>(2x+1)^2 =0`
`<=>2x+1=0`
`<=>x=-1/2`
d)
`9x^2 +6x+1=2`
`<=>(3x+1)^2 =2`
\(< =>\left[{}\begin{matrix}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{\sqrt{2}-1}{3}\\x=\dfrac{-\sqrt{2}-1}{3}\end{matrix}\right.\)
15 tháng 8 2017
\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-2x^3-4x^3+4x^2-3x^2+3x+3x-3=0\)
\(\Leftrightarrow2x^3\left(x-1\right)-4x^2\left(x-1\right)-3x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^3-4x^2-3x+3\right)=0\)
20 tháng 8 2017
Đã có đáp án:
2x^4-6x^3+x^2+6x-3=0
2x^4-6x^3-3x^2-2x^2-6x-3=0
2x^2(x^2-1)-6x(x^2-1)+3(x^2-1)=0
(x^2-1)(2x^2-6x+3)=0
=> { x^2-1=0 =>x=-1;1
Giả phương trình :(*) 2x^2-6x+3=0
4x^2-12x-6=0
(2x)^2-2.2x.3-3=0
(2x-3)^2- (√3)^2=0
( 2x-3)^2=(√3)^2
=> 2x-3=-√3 => 2x= 3-√3 => x=(3-√3)/2
2x-3=√3 => 2x=√3+3 => x=(√3+3)/2
Vậy x....
KH
1
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 12 2020
ĐKXĐ: \(-2\le x\le2\)
\(\Leftrightarrow\sqrt{2x+4}-2\sqrt{2-x}=\dfrac{6x-4}{\sqrt{x^2+4}}\)
\(\Leftrightarrow\dfrac{6x-4}{\sqrt{2x+4}+2\sqrt{2-x}}=\dfrac{6x-4}{\sqrt{x^2+4}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\\sqrt{2x+4}+2\sqrt{2-x}=\sqrt{x^2+4}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+4+8-4x+4\sqrt{2\left(4-x^2\right)}=x^2+4\)
\(\Leftrightarrow4\sqrt{2\left(4-x^2\right)}=x^2+2x-8\)
\(\Leftrightarrow4\sqrt{2\left(4-x^2\right)}=\left(x+4\right)\left(x-2\right)\)
Do \(x\le2\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP\le0\end{matrix}\right.\)
Dấu "=" xảy ra khi và chỉ khi \(x=2\)
Vậy pt có 2 nghiệm \(x=\left\{\dfrac{2}{3};2\right\}\)
\(P=-x^2+6x-4=-\left(x^2-6x+9\right)+5=\left(\sqrt{5}\right)^2-\left(x-3\right)^2.\)
\(=\left(\sqrt{5}-x+3\right)\left(\sqrt{5}+x-3\right)=-\left(x-\left(3+\sqrt{5}\right)\right)\cdot\left(x-\left(3-\sqrt{5}\right)\right)\)
Vậy nghiệm của P là: \(\orbr{\begin{cases}x_1=3+\sqrt{5}\\x_2=3-\sqrt{5}\end{cases}}\)