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a) Ta có: \(x\left(x+3\right)-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
b) Ta có: \(4x^2-1+x\left(2x-1\right)=0\)
\(\Rightarrow\left(2x\right)^2-1^2+x\left(2x-1\right)=0\)
\(\Rightarrow\left[\left(2x\right)^2-1^2\right]+x\left(2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)+x\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1+x\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{-1}{3}\right\}\)
a, x(x+3)-2x-6=0
⇔x(x+3)-2(x+3)=0
⇔(x+3)(x-2)=0
⇔x+3=0 hoặc x-2=0
⇔x= -3 hoặc x=2
b, 4x2-1+x(2x-1)=0
⇔(2x-1)(2x+1)+x(2x-1)=0
⇔(2x-1)(2x+1+x)=0
⇔(2x-1)(3x+1)=0
⇔2x-1=0 hoặc 3x+1=0
⇔x=1/2 hoặc x= -1/3
a) \(\dfrac{2}{x+3}+\dfrac{1}{x}\) MTC: \(x\left(x+3\right)\)
\(=\dfrac{2x}{x\left(x+3\right)}+\dfrac{x+3}{x\left(x+3\right)}\)
\(=\dfrac{2x+x+3}{x\left(x+3\right)}\)
\(=\dfrac{3x+3}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\)
\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\) MTC: \(2\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-2x.2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)-4x}{2\left(x-1\right)}\)
\(=\dfrac{x+1-4x}{2\left(x-1\right)}\)
\(=\dfrac{1-3x}{2\left(x-1\right)}\)
c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}\)
\(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\) MTC: \(6y\left(y-6\right)\)
\(=\dfrac{y\left(y-12\right)}{6y\left(y-6\right)}+\dfrac{6.6}{6y\left(y-6\right)}\)
\(=\dfrac{y\left(y-12\right)+6^2}{6y\left(y-6\right)}\)
\(=\dfrac{y^2-12y+6^2}{6y\left(y-6\right)}\)
\(=\dfrac{\left(y-6\right)^2}{6y\left(y-6\right)}\)
\(=\dfrac{y-6}{6y}\)
Bạn Nguyễn Nam làm sai câu b rồi , làm lại cho tất nè
a) \(\dfrac{2}{x+3}+\dfrac{1}{x}=\dfrac{2x+x+3}{x\left(x+3\right)}=\dfrac{3x+3}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2x+1-4x}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x+1\right)}\)
c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)
\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{\left(y-6\right)^2}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)
d) \(\dfrac{6x}{x+3}+\dfrac{3}{2x+6}=\dfrac{6x}{x+3}+\dfrac{3}{2\left(x+3\right)}=\dfrac{12x}{2\left(x+3\right)}\)( sửa đề )
a) x(y + 4) + 3(y + 4) = 19
<=> (y + 4)(x + 3) = 19
| x + 3 | 19 | 1 | -1 | -19 |
| y + 4 | 1 | 19 | -19 | -1 |
| x | 16 | -2 | -4 | - 22 |
| y | -3 | 15 | -23 | -5 |
pn kẻ bảng ghi y + 4 trước nha, xách hàng 2 lên hàng đầu, hàng 4 lên hàng 3
b) xy + 3x - 2y - 7 = 0
xy + 3x - 2y - 6 = 1
x(y + 3) - 2(y + 3)= 1
(y + 3)(x - 2) = 1
| y + 3 | - 1 | 1 |
| x - 2 | - 1 | 1 |
| y | - 4 | - 2 |
| x | 1 | 3 |
c) 2y(x + 3) = x + 13
2xy + 6y = x + 13
2xy - x + 6y - 3 = 10
x(2y - 1) + 3(2y - 1) = 10
(2y - 1)(x + 3) = 10
tới đây pn lm tương tự như mấy câu kia
d) y(x - 2) + 3x - 6 = 2
y(x - 2) + 3(x - 2) = 2
(x - 2)(y + 3) = 2
| x - 2 | - 1 | - 2 | 1 | 2 |
| y + 3 | -2 | -1 | 2 | 1 |
| x | 1 | 0 | 3 | 4 |
| y | -5 | -4 | -1 | -1 |
e) xy - x + 5y - 7 = 0
xy - x + 5y - 5 = 2
x(y - 1) + 5(y - 1) = 2
(y - 1)(x + 5) = 2
| y - 1 | - 2 | - 1 | 2 | 1 |
| x + 5 | - 1 | - 2 | 1 | 2 |
| y | -1 | 0 | 3 | 2 |
| x | -6 | -7 | -4 | - 3 |
a) x(4x + 2) = 4x2 - 14
⇔ 4x2 + 2x = 4x2 - 14
⇔ 4x2 - 4x2 + 2x = -14
⇔ 2x = -14
⇔ x = -7
Vậy tập nghiệm S = ......
b) (x2 - 9)(2x - 1) = 0
⇔ x2 - 9 = 0 hoặc 2x - 1 = 0
⇔ x2 = 9 hoặc 2x = 1
⇔ x = 3 hoặc -3 hoặc x = \(\dfrac{1}{2}\)
Vậy .......
c) \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{x^2-4}\)
⇔ \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{\left(x-2\right)\left(x+2\right)}\)
ĐKXĐ: x - 2 ≠ 0 và x + 2 ≠ 0
⇔ x ≠ 2 và x ≠ -2MSC (mẫu số chung): (x - 2)(x + 2)Quy đồng mẫu hai vế và khử mẫu ta được:3x + 6 + 4x - 8 = x - 12⇔ 3x + 4x - x = 8 - 6 - 12⇔ 6x = -10⇔ x = \(-\dfrac{5}{3}\) (nhận)Vậy ........
( 2x - 1 ) - x = 0
=> 2x - 1 = x
=> 2x - x = 1
=> x = 1
( x - 1 )( 2x - 3) = 0
=> \(\orbr{\begin{cases}x-1=0\\2x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=\frac{3}{2}\end{cases}}\)
Vậy tập nghiệm của phương trình là S = { 1 ; 3/2 }
\(\frac{x}{x+1}=\frac{x+2}{x-1}\)( đkxđ : \(x\ne\pm1\))
( Chỗ này chưa học kĩ nên chưa hiểu lắm :]
Câu a đề sai
b
\(Q=y^4+64\)
\(Q=\left(y^4+16y^2+64\right)-16y^2\)
\(Q=\left(y^2+8\right)^2-\left(4y\right)^2\)
\(Q=\left(y^2+8-4y\right)\left(y^2+8+4y\right)\)
sai rồi o ra được đâu
x^2 +2x-x-2 =2
<=> x^2 +x-4=0
<=> (x+1/2)^2=9/2
<=> x +1/2 = 3/ căn 2 hoặc -3/ căn 2
<=> x = 3/ căn 2 -1/2 hoặc -3/căn 2 - 1/2
nếu muốn suy ra x(x+1)^2=0 thì đề phải là :
x^3+2x^2-x-2=0