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a: \(=\dfrac{4x^3+8x^2-11x+3-\left(x^2-5\right)\left(2x-1\right)-2x^3-5x^2+x+1}{\left(2x-1\right)^3}\)
\(=\dfrac{2x^3+3x^2-10x+4-2x^3+x^2+10x-5}{\left(2x-1\right)^3}\)
\(=\dfrac{4x^2-1}{\left(2x-1\right)^3}=\dfrac{2x+1}{\left(2x-1\right)^2}\)
b: \(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1+x^{32}}\)
12: \(\Leftrightarrow\left(\dfrac{4x^2+16-3x^2-18}{x^2+6}\right)=\left(\dfrac{3}{x^2+1}-1\right)+\left(\dfrac{5}{x^2+3}-1\right)+\left(\dfrac{7}{x^2+5}-1\right)\)
=>x2-2=0
hay \(x=\pm\sqrt{2}\)
13: \(\Leftrightarrow x^5-x^4-x^3-x^2-x-2=0\)
\(\Leftrightarrow x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2=0\)
=>x-2=0
=>x=2
điều kiện xác định \(x\ne0\)
ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-\left(x^3+2x^2+4x-2x^2-4x-8\right)}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-x^3-2x^2-4x+2x^2+4x+8}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{-x^2+2x+12}{x^4+4x^2+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)
\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)
tới đây bn bấm máy tính nha
ĐKXĐ: x∉{0;2}
Ta có: \(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
\(\Leftrightarrow\frac{5-x}{4x\left(x-2\right)}+\frac{7}{8x}-\frac{x-1}{2x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}-\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}=0\)
Suy ra: \(10-2x+7x-14-4x+4-x=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;2\right\}\end{matrix}\right.\)
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
a) ĐKXĐ: \(x\notin\left\{-3;2;-1;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x-1}\)
\(\Leftrightarrow\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{2}{\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x-1}\)
\(\Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}-\dfrac{2\left(x-2\right)}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-3}{2x-1}\)
\(\Leftrightarrow\dfrac{5x+5-2x+4}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-3}{2x-1}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{3}{1-2x}\)
\(\Leftrightarrow\dfrac{3\left(x+3\right)}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{3}{1-2x}\)
\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x-2\right)}=\dfrac{3}{1-2x}\)
Suy ra: \(\left(x+1\right)\left(x-2\right)=1-2x\)
\(\Leftrightarrow x^2-x-2-1+2x=0\)
\(\Leftrightarrow x^2+x-3=0\)
\(\Delta=1^2-4\cdot1\cdot\left(-3\right)=13\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{13}}{2}\left(nhận\right)\\x_2=\dfrac{-1+\sqrt{13}}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{-1-\sqrt{13}}{2};\dfrac{-1+\sqrt{13}}{2}\right\}\)
Lớp 8 nên chưa học biệt thức delta
Ta có: \(x^2+x-3=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{13}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{13}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{13}-1}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\) \(\dfrac{7}{8x}\)+\(\dfrac{5-x}{4x\left(x-2\right)}\)= \(\dfrac{x-1}{2x\left(x-2\right)}\)+ \(\dfrac{1}{8\left(x-2\right)}\)
\(\Rightarrow\) 7(x-2) + 2(5-x) = 4(x-1) +x
\(\Leftrightarrow\) 7x-2x+10-2x= 4x-4+x
\(\Leftrightarrow\)7x-2x-2x-4x-x = -4-10
\(\Leftrightarrow\) -2x = -14
\(\Leftrightarrow\) x = 7
Vậy phương trình có nghiệm x=7
⇔ 78x78x+5−x4x(x−2)5−x4x(x−2)= x−12x(x−2)x−12x(x−2)+ 18(x−2)18(x−2)
⇔ 7(x-2)8x(x-2)78x+2(5−x)8x(x−2)5−x4x(x−2)= 4(x−1)28x(x−2)x−12x(x−2)+ x8x(x−2)
18(x−2)
⇒7(x-2)+2(5-x)=4(x-1)+x
⇔ 7x-2x+10-2x= 4x-4+x
⇔7x-2x-2x-4x-x = -4-10
⇔ -2x = -14
⇔ x = 7
vậy tập của phương trình là: S=7}
Đề bài sai, pt này ko giải được
Đề đúng: \(\dfrac{4x^2+16}{x^2+6}=...\)
Mẫu số bên trái thừa mất số 1