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Giải:
a) \(\frac{3x+2}{3x-2}\)−62+3x=9x29x2−4 ⇔ \(\frac{9x^2+12x+4}{\left(3x-2\right)\left(3x+2\right)}\) - \(\frac{18x-12}{\left(3x-2\right)\left(3x+2\right)}\) = \(\frac{9x^2}{9x^2-4}\) ⇔ 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!
a,\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)
ĐKXĐ: x≠1/4, x≠-1/4
⇔\(-\frac{3}{4x-1}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)
⇔\(\frac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\frac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\frac{3+6x}{16x^2-1}\)
⇒-12x-3=8x-2-3-6x
⇔8x-6x+12x=-3+2+3
⇔14x=2
⇔x=1/7(tmđk)
Vậy phương trình có nghiệm là x=1/7
b, \(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\) (2)
ĐKXĐ: x≠0, x≠2
(2)⇔\(\frac{2\left(5-x\right)}{2.4x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4.\left(x-1\right)}{4.2x\left(x-2\right)}+\frac{x}{8.x\left(x-2\right)}\)
⇒10-2x+7x-14=4x-4+x
⇔-2x+7x-4x-x=-4-10+14
⇔0x=0
⇔ x∈R
Vậy phương trình có nghiệm là x∈R và x≠0, x≠2
c, \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) (3)
ĐKXĐ: x≠0
(3)⇒x(x+1)(x2-x+1)-x(x-1)(x2+x+1)=3
⇔x4+x-x4+x=3
⇔2x=3
⇔x=3/2(tmđk)
Vậy phương trình có nghiệm là x=3/2
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
a) Đề ( \(x\ne\pm1\))
>\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{4}{\left(x+1\right)\left(x-1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=4\\ \Leftrightarrow2.2x=4\Leftrightarrow x=1\left(kothỏa\right)\)
Vậy \(S=\varnothing\)
b) đề \(\left(x\ne-\frac{1}{2},\frac{1}{2}\right)\)
\(\frac{32x^2}{12\left(1-2x\right)\left(1+2x\right)}=\frac{-8x\left(1+2x\right)}{12\left(1-2x\right)\left(1+2x\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(1+2x\right)}\\ \Leftrightarrow32x^2=-8x-16x^2-3-12x+48x^2\\ \Leftrightarrow20x+3=0\Leftrightarrow x=\frac{20}{3}\left(thỏadk\right)\)
Vậy \(S=\left\{\frac{20}{3}\right\}\)
b, \(\frac{1}{x-1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\left(ĐKXĐ:x\ne\pm1;x\ne2\right)\)
\(\Leftrightarrow\)\(\frac{1}{x-1}+\frac{5}{2-x}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(2-x\right)+5\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(2-x\right)\left(x-1\right)}=\frac{15\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(2-x\right)}\)
Suy ra:
\(\Leftrightarrow\)(x+1)(2-x)+5(x-1)(x+1) = 15(x-1)
\(\Leftrightarrow\)2x-x2-x+2+5x2-5 = 15x-15
\(\Leftrightarrow\)2x-x2-x+5x2-15x = -15+5-2
\(\Leftrightarrow\)4x2-14x = -12
\(\Leftrightarrow4x^2-14x+12=0\)
\(\Leftrightarrow4x^2-8x-6x+12=0\)
\(\Leftrightarrow\)4x(x-2) - 6(x-2) = 0
\(\Leftrightarrow\left(x-2\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(kotm\right)\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất x = \(\frac{3}{2}\)
b/ \(3-100x+8x^2=8x^2+x-300\)
\(\Leftrightarrow-101x=-303\)
\(\Rightarrow x=3\)
c/ \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-79x=-158\)
\(\Rightarrow x=2\)
d/ \(3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
\(\Leftrightarrow-6x=5\)
\(\Rightarrow x=-\frac{5}{6}\)
e/ \(30x-6\left(2x-5\right)+5\left(x+8\right)=210+10\left(x-1\right)\)
\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)
\(\Leftrightarrow13x=130\)
\(\Rightarrow x=10\)
\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)
\(\Rightarrow A_{min}=-3\) khi \(x=2\)
\(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)
\(\Rightarrow B_{min}=10\) khi \(x=-\frac{1}{2}\)
\(C=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow C_{min}=-36\) khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(D=-x^2-8x-16+21=21-\left(x+4\right)^2\le21\)
\(\Rightarrow C_{max}=21\) khi \(x=-4\)
\(E=-x^2+4x-4+5=5-\left(x-2\right)^2\le5\)
\(\Rightarrow E_{max}=5\) khi \(x=2\)
\(\frac{x+3}{x^2+x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)+1\)
ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\frac{6}{\left(x-2\right)\left(x+2\right)}+1\)
\(=\frac{\left(x+3\right)\left(x-2\right)\left(x+2\right)}{6\left(x^2+x+6\right)}+1\)
\(=\frac{x^3+3x^2-4x-12}{6x^2+6x+36}+\frac{6x^2+6x+36}{6x^2+6x+36}\)
\(=\frac{x^3+3x^2-4x-12+6x^2+6x+36}{6x^2+3x+36}\)
\(=\frac{x^3+9x^2+2x+24}{6x^2+3x+36}\)
ơ lag hai dòng cuối :v
bạn sửa mẫu thức ở hai dòng cuối thành 6x2 + 6x + 36 dùm mình nhé -.- dạo này lú quá