giải phương trình :
\(\dfrac{80}{x+4}\)+ \(\dfrac{80}{x-4}\) = \(\dfrac{25}{3}\)
giải các phương trình sau
1, \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
2, \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)
3, \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)
Suy ra: \(5x^2+3x-9=5x^2-5x\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(tm\right)\)
2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)
\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(6x=3x-15\)
\(\Leftrightarrow3x=-15\)
hay \(x=-5\left(loại\right)\)
2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)
Vậy pt vô nghiệm.
3. ĐKXĐ: $x\neq \pm 4$
PT \(\Leftrightarrow \frac{-3(x+4)}{(x-4)(x+4)}-\frac{3-5x}{(x-4)(x+4)}=\frac{x-4}{(x-4)(x+4)}\)
\(\Rightarrow -3(x+4)-(3-5x)=x-4\)
\(\Leftrightarrow 2x-15=x-4\Leftrightarrow x=11\) (thỏa mãn)
Giải các phương trình sau: (TM ĐK)
1) \(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\)
2) \(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
3) \(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{x-5}{2x^2+10}\)
4) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
5) \(\left(1-\dfrac{x-1}{x+1}\right)\left(x+2\right)=\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\)
mng giúp mk bài này nha. Cảm ơn bạn nhiều
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
Giải phương trình: \(x-\dfrac{\dfrac{x}{3}-\dfrac{3+x}{2}}{4}=\dfrac{x-\dfrac{15-7x}{3}}{4}-2x+3\)
\(x-\dfrac{\dfrac{x}{3}-\dfrac{3+x}{2}}{4}=\dfrac{x-\dfrac{15-7x}{3}}{4}-2x+3\)
\(<=>4x-\dfrac{x}{3}+\dfrac{3+x}{2}=x-\dfrac{15-7x}{3}-8x+12\)
`<=>24x-2x+3(3+x)=6x-2(15-7x)-48x+72`
`<=>24x-2x+9+3x=6x-30+14x-48x+72`
`<=>53x=33`
`<=>x=33/53`
Giải phương trình
\(\dfrac{1}{3}\sqrt{9x+9}-2\sqrt{x+1}+8\sqrt{\dfrac{4x+4}{25}}=11\)
\(\dfrac{1}{3}\sqrt[]{9x+9}-2\sqrt[]{x+1}+8\sqrt[]{\dfrac{4x+4}{25}}=11\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt[]{9\left(x+1\right)}-2\sqrt[]{x+1}+8\sqrt[]{\dfrac{4\left(x+1\right)}{25}}=11\)
\(\Leftrightarrow\sqrt[]{x+1}-2\sqrt[]{x+1}+\dfrac{16}{5}\sqrt[]{x+1}=11\)
\(\Leftrightarrow\dfrac{11}{5}\sqrt[]{x+1}=11\)
\(\Leftrightarrow\sqrt[]{x+1}=5\)
\(\Leftrightarrow x+1=25\)
\(\Leftrightarrow x=24\)
Giải các phương trình:
\(\dfrac{x+24}{1996}+\dfrac{x+25}{1995}+\dfrac{x+26}{1994}+\dfrac{x+27}{1993}+\dfrac{x+2036}{4}=0\)
Mỗi số hạng của vế trái cộng thêm 1, vế phải = 5. Mỗi số hạng vế trái có mẫu số giống nhau, bạn đặt x+ 2020 làm nhân tử chung, phần còn lại tự làm nhé.
mấy bài còn lại bạn đăng cx làm tương tự
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
\(\Leftrightarrow\left(\frac{x+24}{1996}+1\right)+\left(\frac{x+25}{1995}+1\right)+\left(\frac{x+26}{1994}+1\right)+\left(\frac{x+27}{1993}+1\right)+\left(\frac{x+2036}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy ....
giải các phương trình sau
a, 4x- 2(1-x)= 5(x-4)
b, \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
c, \(\left(x+2\right)^2-3\left(x+2\right)=0\)
d,\(\dfrac{x-5}{x}+\dfrac{x-3}{x+5}=\dfrac{x-25}{x\left(x+5\right)}\)
a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)
\(\Leftrightarrow4x-2+2x=5x-20\)
\(\Leftrightarrow x=-18\)
b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-3x=-1\)
hay \(x=\dfrac{1}{3}\)
c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
d: Ta có: \(\dfrac{x-5}{x}+\dfrac{x-3}{x+5}=\dfrac{x-25}{x\left(x+5\right)}\)
\(\Leftrightarrow x^2-25+x^2-3x=x-25\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
giải các phương trình sau
1, \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
2, \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
3, \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(3x+9+4x-12=3x-7\)
\(\Leftrightarrow4x=-7+12-9=-4\)
hay \(x=-1\left(nhận\right)\)
2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(3x+12-4x+16=3x-4\)
\(\Leftrightarrow28-4x=-4\)
\(\Leftrightarrow4x=32\)
hay \(x=8\left(tm\right)\)
3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
Suy ra: \(5x^2-12+3x+3=5x^2-5x\)
\(\Leftrightarrow3x-9+5x=0\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(nhận\right)\)
Giải các phương trình sau:
a. \(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
b. \(\dfrac{2}{x^2+4x+3}+\dfrac{5}{x^2+11x+24}+\dfrac{2}{x^2+18x+80}=\dfrac{9}{25}\)
a.
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)=18\)
Đặt \(t=x^2+2x+1=\left(x+1\right)^2\left(t\ge0\right)\)
\(\Rightarrow\left(4t-1\right)\cdot t=18\)
\(\Leftrightarrow\left(2t\right)^2-2\cdot2t\cdot\dfrac{1}{4}+\dfrac{1}{16}=\dfrac{289}{16}\)
\(\Leftrightarrow\left(2t-\dfrac{1}{4}\right)^2=\dfrac{289}{16}\Leftrightarrow\left(t-\dfrac{1}{8}\right)^2=\dfrac{289}{64}\)
\(\Leftrightarrow\left[{}\begin{matrix}t-\dfrac{1}{8}=\dfrac{17}{8}\\t-\dfrac{1}{8}=-\dfrac{17}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\\t=-2\left(loai\right)\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2=\dfrac{9}{4}\Leftrightarrow\left[{}\begin{matrix}x+1=\dfrac{3}{2}\\x+1=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{5}{2};\dfrac{1}{2}\right\}\)
b.
Ta có:
- \(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
- \(x^2+11x+24=x^2+3x+8x+24=x\left(x+3\right)+8\left(x+3\right)=\left(x+3\right)\left(x+8\right)\)
- \(x^2+18x+80=x^2+8x+10x+80=x\left(x+8\right)+10\left(x+8\right)=\left(x+8\right)\left(x+10\right)\)
Thay vào phương trình, ta được:
\(\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{x+10-\left(x+1\right)}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{25}\Leftrightarrow\dfrac{9}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{25}\)
\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=25\)
\(\Leftrightarrow x^2+11x+\dfrac{121}{4}=\dfrac{181}{4}\)
\(\Leftrightarrow\left(x+\dfrac{11}{2}\right)^2=\dfrac{181}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{11}{2}=\dfrac{\sqrt{181}}{2}\\x+\dfrac{11}{2}=-\dfrac{\sqrt{181}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{181}}{2}\\x=\dfrac{-11-\sqrt{181}}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{-11+\sqrt{181}}{2};\dfrac{-11-\sqrt{181}}{2}\right\}\)
2) Tính: (Giải chi tiết từng bước)
a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)
b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
3) Tìm x, biết:
a) \(\sqrt{\left(x-1\right)^2}=4\)
b) \(\sqrt{36x^2-60x+25}=4\)
Bài 2:
a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)
\(=2\sqrt{5^2\cdot5}+\dfrac{3}{2}\sqrt{4^2\cdot5}-\sqrt{6^2\cdot5}-\dfrac{2}{7}\sqrt{7^2\cdot5}\)
\(=10\sqrt{5}+\dfrac{3\cdot4}{2}\sqrt{5}-6\sqrt{5}-\dfrac{2\cdot7}{7}\sqrt{5}\)
\(=10\sqrt{5}+6\sqrt{6}-6\sqrt{5}-2\sqrt{5}\)
\(=8\sqrt{5}\)
b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2-2\cdot2\cdot\sqrt{7}+2^2}-\sqrt{\left(\sqrt{7}\right)^2+2\cdot3\cdot\sqrt{7}+3^2}\)
\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(\sqrt{7}+3\right)^2}\)
\(=\sqrt{7}-2-\sqrt{7}-3\)
\(=-5\)
\(2a,\\ 2\sqrt{125}+\dfrac{3}{2}.\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\\ =2\sqrt{5^2.5}+\dfrac{3}{2}.\sqrt{4^2.5}-\sqrt{6^2.5}-\dfrac{2}{7}.\sqrt{7^2.5}\\ =2.5.\sqrt{5}+\dfrac{3}{2}.4.\sqrt{5}-6\sqrt{5}-\dfrac{2}{7}.7\sqrt{5}\\ =10\sqrt{5}+6\sqrt{5}-6\sqrt{5}-2\sqrt{5}=8\sqrt{5}\)
3:
a: =>|x-1|=4
=>x-1=4 hoặc x-1=-4
=>x=-3 hoặc x=5
b: =>|6x-5|=4
=>6x-5=4 hoặc 6x-5=-4
=>6x=1 hoặc 6x=9
=>x=1/6 hoặc x=3/2