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 \(x=\left\{1;1\pm\sqrt{6}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{17}}{2}< x< 1-\sqrt{6}\\1< x< \frac{-1+\sqrt{17}}{2}\\x>1+\sqrt{6}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{17}}{2}\\1-\sqrt{6}< x< 1\\\frac{-1+\sqrt{17}}{2}< x< 1+\sqrt{6}\end{matrix}\right.\)
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
Giải các phương trình sau:
a)\(\frac{\left(9x-0.7\right)}{4}-\frac{\left(5x-1.5\right)}{7}=\frac{\left(7x-1.1\right)}{3}-\frac{5\left(0.4-2x\right)}{6}\)
b)\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
c)\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}\)
d)\(\frac{8x^2}{3\left(1-4x\right)^2}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
giải các hệ BPT sau:
a) \(\left\{{}\begin{matrix}5x-2>4x+5\\5x-4< x+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\)
g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\)
h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)
j) \(\left\{{}\begin{matrix}\frac{3x+1}{2}-\frac{3-x}{3}\le\frac{x+1}{4}-\frac{2x-1}{3}\\3-\frac{2x+1}{5}>x+\frac{4}{3}\end{matrix}\right.\)
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 :)
Giải các phương trình,bất phương trình:
c,\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
d,\(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\)
e,\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}=0\)
g,\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
h,\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
i,\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
k,\(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)
l,\(\left(x^2-2x+1\right)-4=0\)
m,\(4x^2+4x++1=x^2\)
Xin đáy ai giúp mình đi
bài 1 giải phương trình
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\frac{3}{5x-1}+\frac{3}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8+6x}{16x^2-1}\)
\(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
Giải:
a) ⇔⇔ 9x2 + 12x + 4 - 18x + 12 = 9x2 ⇔ 9x2 + 12x + 4 - 18x + 12 - 9x2 = 0
⇔ 16 + 6x = 0 ⇔ 2(8 + 3x) = 0 ⇔ 8 + 3x = 0 ⇔ x = \(\frac{-8}{3}\)
Vậy nghiệm của phương trình là x = \(\frac{-8}{3}\) .
b) \(\frac{3}{5x-1}+\frac{3}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\text{⇔ }\frac{-3}{1-5x}+\frac{-3}{5x-3}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
⇔ \(\frac{9-15x}{\left(1-5x\right)\left(5x-3\right)}+\frac{15x-3}{\left(1-5x\right)\left(5x-3\right)}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\) ⇔ 9 - 15x + 15x - 3 = 4
⇔ 8 = 4 ( vô lí)
Vậy phương trình trên vô nghiệm.
Mình chỉ làm 2 câu a, b thôi nhé! Các bài tập này cách làm giống nhau, bạn tự hoàn thành những bài còn lại nhé!
Giải phương trình :
\(\left(\frac{8}{3}\right)^{x^2-x+1}\left(\frac{3}{5}\right)^{2x^2-3x+2}\left(\frac{5}{7}\right)^{3x^2-4x+3}\left(\frac{7}{2}\right)^{4x^2-5x+4}=210^{\left(x-1\right)^2}\)
\(\Leftrightarrow\frac{2^{3x^2-3x+1}}{3^{x^2-x+1}}.\frac{3^{2x^2-3x+2}}{5^{2x^2-3x+2}}.\frac{5^{3x^2-4x+3}}{7^{3x^2-4x+3}}.\frac{7^{4x^2-5x+4}}{2^{4x^2-5x+4}}=210^{\left(x-1\right)^2}\)
\(\Leftrightarrow\frac{\left(3.5.7\right)^{x^2-x+1}}{2^{x^2-2x+1}}=2^{\left(x-1\right)^2}.\left(3.5.7\right)^{\left(x-1\right)^2}\)
\(\Leftrightarrow105^x=2^{2\left(x-1\right)^2}\)
Lấy Logarit cơ số 2 hai vế, ta được :
\(2\left(x-1\right)^2=\left(\log_2105\right)x\)
\(\Leftrightarrow2x^2-\left(4+\log_2105\right)x+2=0\)
\(\Leftrightarrow x=\frac{\left(2+\log_2105\right)\pm\sqrt{\log^2_2105+8\log_2105}}{4}\)
Vậy phương trình đã cho có 2 nghiệm
Bài 1:Giải phương trình
a)\(10x^2-5x\left(2x+3\right)=15\)
b)\(3x-7-\left(3-4x\right)\left(2x+1\right)=4x\left(2x-7\right)\)
c)\(\left(4x-5\right)^2-\left(7-2x\right)=4\left(2x-4\right)^2+6x\)
Bài 2:Giải phương trình
a)\(\frac{3\left(x-1\right)}{2}+4=\frac{2x}{3}+\frac{4-5x}{6}\)
b)\(\frac{4-x}{7}-\frac{1}{7}\left(\frac{7+3x}{9}+\frac{5-2x}{2}\right)=4-\frac{4x}{3}\)
c)\(\frac{2}{9}\left(2x-5\right)-\frac{5}{3}\left[\left(x-2\right)-\frac{7}{12}\right]=\frac{3}{4}\left(x-3\right)\)
Bài 3:Giải phương trình
a)\(\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\)
b)\(2x\left(x-3\right)+5\left(x-3\right)=0\)
c)\(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
Bài 4:Tìm m để phương trình sau có nghiệm bằng 7:\(\left(2m-5\right)x-2m^2+8=43\)
Bài 5:Giải phương trình
a)\(\left(2x-1\right)^2-\left(2x+1\right)^2=0\)
b)\(\frac{1}{27}\left(x-3\right)^3-\frac{1}{125}\left(x-5\right)^3=0\)
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
Bài 4 xem lại đề nhé bác
