GPT:
\(\left(1\right)\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\left(2\right)\frac{x-3}{x+3}-\frac{x+3}{x-3}=-\frac{5}{x^2-9}\)
Những câu hỏi liên quan
1)2x(25x-4)-(5x-2)(5x+1)8 / 5)2left(x-2right)-3left(3x-1right)left(x-3right)
2)x(4x-3)-(2x-2)(2x-1)5 / 6)frac{2}{x+1}-frac{1}{x-2}frac{3x-11}{left(x+1right)left(x-2right)}
3)frac{5}{2x+3}+frac{3}{9-x^2}frac{8}{7left(x3right)} / 7)frac{5x-2}{6}+frac{3-4x}{2}2-frac{x+7}{3}
4)frac{2}{3left(x-2right)}+frac{5}{12-3x^2}frac{3}{4left(x+2right)} /...
Đọc tiếp
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
Đúng 0
Bình luận (0)
Bài 3 : Xét dấu biểu thức sau :
1 , fleft(xright)frac{x-7}{4x^2-19x+12}
2 , fleft(xright)frac{11x+3}{-x^2+5x-7}
3 , fleft(xright)frac{3x-2}{x^3-3x^2+2}
4 , fleft(xright)frac{x^2+4x-12}{sqrt{6}x^2+3x+sqrt{2}}
5 , fleft(xright)frac{x^2-3x-2}{-x^2+x-1}
6 , fleft(xright)frac{x^3-5x+4}{x^4-4x^3+8x-5}
7 , fleft(xright)frac{left(x+3right)left(x-2right)left(-2x^2+x-1right)}{left(2x-5right)left(x^2+3x-10right)}
8 , fleft(xright)left(-x^2+x-1right)left(6x^2-5x+1right)
9 , fleft(xright)frac{x^2-...
Đọc tiếp
Bài 3 : Xét dấu biểu thức sau :
1 , \(f\left(x\right)=\frac{x-7}{4x^2-19x+12}\)
2 , \(f\left(x\right)=\frac{11x+3}{-x^2+5x-7}\)
3 , \(f\left(x\right)=\frac{3x-2}{x^3-3x^2+2}\)
4 , \(f\left(x\right)=\frac{x^2+4x-12}{\sqrt{6}x^2+3x+\sqrt{2}}\)
5 , \(f\left(x\right)=\frac{x^2-3x-2}{-x^2+x-1}\)
6 , \(f\left(x\right)=\frac{x^3-5x+4}{x^4-4x^3+8x-5}\)
7 , \(f\left(x\right)=\frac{\left(x+3\right)\left(x-2\right)\left(-2x^2+x-1\right)}{\left(2x-5\right)\left(x^2+3x-10\right)}\)
8 , \(f\left(x\right)=\left(-x^2+x-1\right)\left(6x^2-5x+1\right)\)
9 , \(f\left(x\right)=\frac{x^2-x-2}{-x^2+3x+4}\)
10 , \(f\left(x\right)=\left(x^2-5x+4\right)\left(2-5x+2x^2\right)\)
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
5.
\(f\left(x\right)=\frac{x^2-3x-2}{-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\frac{3\pm\sqrt{17}}{2}\)
\(f\left(x\right)>0\Rightarrow\frac{3-\sqrt{17}}{2}< x< \frac{3+\sqrt{17}}{2}\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3-\sqrt{17}}{2}\\x>\frac{3+\sqrt{17}}{2}\end{matrix}\right.\)
6.
\(f\left(x\right)=\frac{\left(x-1\right)\left(x^2+x-4\right)}{\left(x-1\right)^2\left(x^2-2x-5\right)}=\frac{x^2+x-4}{\left(x-1\right)\left(x^2-2x-5\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{6}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{17}}{2}< x< 1-\sqrt{6}\\1< x< \frac{-1+\sqrt{17}}{2}\\x>1+\sqrt{6}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{17}}{2}\\1-\sqrt{6}< x< 1\\\frac{-1+\sqrt{17}}{2}< x< 1+\sqrt{6}\end{matrix}\right.\)
Xem thêm câu trả lời
Mọi người giải giúp mình bài này với:B2: Giải các PT sau:l) left(x-1right)left(5x+3right)left(3x-8right)left(x-1right)m) 2xleft(x-1right)x^2-1n) left(2-3xright)left(x+11right)left(3x-2right)left(2-5xright)o) frac{3}{7}x-1frac{1}{7}xleft(3x-7right)p) left(x-frac{3}{4}right)^2+left(x-frac{3}{4}right)left(x-frac{1}{2}right)0q) frac{1}{x}x+2left(frac{1}{x}+2right)left(x^2+1right)r) left(2x+3right)left(frac{3x+8}{2-7x}+1right)left(x-5right)left(frac{3x+8}{2-7x}+1right)s) left(x+2right)left(x-3right)l...
Đọc tiếp
Mọi người giải giúp mình bài này với:
B2: Giải các PT sau:
l) \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
m) \(2x\left(x-1\right)=x^2-1\)
n) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
o) \(\frac{3}{7}x-1=\frac{1}{7}x\left(3x-7\right)\)
p) \(\left(x-\frac{3}{4}\right)^2+\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}\right)=0\)
q) \(\frac{1}{x}x+2=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
r) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
s) \(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)
Các Pro giúp mình với !
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
Xem thêm câu trả lời
1 . left(3x-2right)left(x^2-1right)left(9x^2-4right)left(x+1right)2 . frac{7}{8}x-5left(x-9right)frac{20x+1,5}{6}3 . 4left(0,5-1,5xright)frac{5x-6}{3}4 . left(3x-1right)left(x^2+2right)left(3x-1right)left(7x-10right)5 . frac{8x^2}{3left(1-4x^2right)}+frac{1}{2x+7}frac{6}{x^2-9}
Đọc tiếp
1 . \(\left(3x-2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
2 . \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
3 . \(4\left(0,5-1,5x\right)=\frac{5x-6}{3}\)
4 . \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
5 . \(\frac{8x^2}{3\left(1-4x^2\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Mấy câu như vậy bạn tải photomath về dùng nhé :)
1. Thực hiện phép tínha) left(3x^2y-6xy+9xright).left(frac{-4}{3xy}right)b) left(frac{1}{3}x+2yright).left(frac{1}{9}x^2-frac{2}{3}xy+4y^2right)c) (x-2) ( x^2-5x+1) + x ( x^2+11)d) (x - 3y ) left(x^2+3xy+9y^2right)e) left(3+xright)left(x^2+3x-5right)f) (x+2)(x-2)-(2x+1)2. Rút gọn biểu thứca ) left(3x+2right)^2+2left(2+3xright)left(1-2yright)+left(2y-1right)^2b ) left(3x+1right)^2-2left(3x+1right)left(3x+5right)+left(3x+5right)^2c) 2xleft(2x-1right)^2-3xleft(x+3right)left(x-3right)-4xleft(x+1righ...
Đọc tiếp
1. Thực hiện phép tính
a) \(\left(3x^2y-6xy+9x\right).\left(\frac{-4}{3xy}\right)\)
b) \(\left(\frac{1}{3}x+2y\right).\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)
c) (x-2) ( \(x^2-5x+1\)) + x ( \(x^2+11\))
d) (x - 3y ) \(\left(x^2+3xy+9y^2\right)\)
e) \(\left(3+x\right)\left(x^2+3x-5\right)\)
f) (x+2)(x-2)-(2x+1)
2. Rút gọn biểu thức
a ) \(\left(3x+2\right)^2+2\left(2+3x\right)\left(1-2y\right)+\left(2y-1\right)^2\)
b ) \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
c) \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
cau a : (3x^2y-6xy+9x)(-4/3xy)
=-4/3xy.3x^2y+4/3xy.6xy-4/3xy.9x
=-4x+8-8y
cau b : (1/3x+2y)(1/9x^2-2/3xy+4y^2)
=(1/3)^3-2/9x^2y+8y^3+4/3xy^2+2/9x^2y-4/3xy^2+8y^3
=(1/3)^3 + (2y)^3x-2
cau c : (x-2)(x^2-5x+1)+x(x^2+11)
=x^3-5x^2+x-2x^2+10x-2+x^3+11x
=2x^3-7x^2+22x-2
cau d := x^3 + 6xy^2 -27y^3
cau e := x^3 + 3x^2 -5x - 3x^2y - 9xy = 15y
cau f := x^2-2x+2x -4-2x-1
= x(x-2)-5
cau e la + 15y ko phai =15y
Giải các PT sau :
a,frac{1-6x}{x-2}+frac{9x+4}{x+2}frac{xleft(3x-2right)+1}{left(x+2right)left(x-2right)}
b,(2 - 3x) (x + 11) (3x - 2) (2 - 5x)
c,frac{3x-2}{6}-5frac{3-2left(x+7right)}{4}
d,frac{x}{x-3}+frac{x}{x
+1}frac{2x}{left(x+1right)left(x-3right)}
e,frac{x+1}{5}+frac{x+2}{4}frac{x+3}{3}+frac{x+4}{2}
Đọc tiếp
Giải các PT sau :
a,\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{\left(x+2\right)\left(x-2\right)}\)
b,(2 - 3x) (x + 11) = (3x - 2) (2 - 5x)
c,\(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\)
d,\(\frac{x}{x-3}+\frac{x}{x +1}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
e,\(\frac{x+1}{5}+\frac{x+2}{4}=\frac{x+3}{3}+\frac{x+4}{2}\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\\ \left(\frac{3x}{1-3n}+\frac{2n}{3x+1}\right):\left(\frac{6x^2+10x}{1-6x+9x^2}\right)\\ \left(\frac{9}{x^3-9n}+\frac{1}{x+3}\right):\left(\frac{x}{3n+9}\right)\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
Đúng 0
Bình luận (0)
1. frac{1}{2}x^2-left(frac{1}{2}x-4right)frac{1}{2}x-14
2. 3left(1-4xright)left(x-1right)+4left(3x-2right)left(x+3right)-27
3. 6xleft(5x+3right)+3xleft(1-10xright)7
4. left(3x-3right)left(5-21xright)+left(7x+4right)left(9x-5right)44
5. left(-2+x^3right)left(-2+x^2right)left(-2+x^2right)1
Đọc tiếp
1. \(\frac{1}{2}x^2-\left(\frac{1}{2}x-4\right)\frac{1}{2}x=-14\)
2. \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)
3. \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
4. \(\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\)
5. \(\left(-2+x^3\right)\left(-2+x^2\right)\left(-2+x^2\right)=1\)
Yêu cầu đề bài là gì hả bạn?
Giải các phương trình
a. \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
b. \(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)
c. \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
d.\(\left(2x^2+1\right)\left(4x-3\right)=\left(2x^2+1\right)\left(x-12\right)\)
a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
\(=>\frac{1-x+x+1}{x+1}+2=\frac{1}{x+1}+2\)
\(=>\frac{2}{x+1}=\frac{1}{x+1}\)
\(=>2x+2=x+1\)
\(=>2x-x=1-2=-1\)
\(=>x=-1\)
vậy nghiệm của phương trình trên là {-1}
À quên ĐKXĐ của câu a là \(x\ne-1\)
Nên \(x\in\varnothing\)nhé :v
\(\left(2x^2+1\right)\left(4x-2\right)=\left(2x^2+1\right)\left(x-12\right)\)
\(\Leftrightarrow8x^3-6x^2+4x-3=2x^3-24x^2+x-12\)
\(\Leftrightarrow8x^3-6x^2+4x-3-2x^3+24x^4-x+12=0\)
\(\Leftrightarrow6x^3+18x^2+3x+9=0\)
\(\Leftrightarrow3\left(x+3\right)\left(2x^2+1\right)=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)