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\(\frac{36}{x+6}+\frac{36}{x-6}=\) \(4,5\)\(\left(ĐKCĐ:x\ne\pm6\right)\)
\(\Leftrightarrow\frac{36\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}+\frac{36\left(x+6\right)}{\left(x+6\right)\left(x-6\right)}\)\(=\frac{4,5\left(x-6\right)\left(x+6\right)}{\left(x-6\right)\left(x+6\right)}\)
\(\Leftrightarrow\frac{36x-216}{\left(x-6\right)\left(x+6\right)}+\frac{36x+216}{\left(x-6\right)\left(x+6\right)}\)\(=\frac{4,5x^2-162}{\left(x-6\right)\left(x+6\right)}\)
\(\Rightarrow36x-216+36x+216=4,5x^2-162\)
( đến đây giải phương trình ra rồi đối chiếu đkxđ là xong )
\(\frac{36}{x+6}+\frac{36}{x-6}=4,5\)
\(\frac{36}{x+6}+\frac{36}{x-6}=\frac{4,5\left(x+6\right)\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}\)
\(DKXD:\hept{\begin{cases}x+6\ne0\\x-6\ne0\\\left(x+6\right)\left(x-6\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-6\\x\ne6\end{cases}}\)
\(\frac{72x}{\left(x+6\right)\left(x-6\right)}=\frac{4,5\left(x+6\right)\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}\)
\(4,5x^2+72x-162=0\)
\(4,5x^2-9x+81x-162=0\)
\(4,5\left(x-2\right)+81\left(x-2\right)=0\)
\(\left(x-2\right)\left(4,5x-81\right)=0\)
\(\left(x-2\right)4,5\left(x-18\right)=0\)
\(\hept{\begin{cases}x-2=0\\x-18=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\x=18\end{cases}}\)
\(\dfrac{90}{x}-\dfrac{36}{x-6}=2\) ( x # 0 ; x # 6)
⇔ \(\dfrac{90\left(x-6\right)-36x}{x\left(x-6\right)}=\dfrac{2x\left(x-6\right)}{x\left(x-6\right)}\)
⇔ 90x - 540 - 36x = 2x2 - 12x
⇔-2x2 + 66x - 540 = 0
⇔ -2( x2 - 33x +270 ) = 0
⇔ x2 - 18x - 15x + 270 = 0
⇔ x( x - 18) - 15( x - 18) = 0
⇔ ( x - 18)( x - 15) = 0
⇔ x = 18 ( TM) hoac x = 15 ( TM)
KL........
ta có:
(\(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\)):\(\dfrac{2x-6}{x^2+6x}\)+\(\dfrac{x}{6-x}\)
= (\(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)}{\left(x+6\right)\left(x-6\right)}\)):\(\dfrac{2x-6}{x^2+6x}\)+\(\dfrac{x}{6-x}\)
= (\(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\)).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{12x-36}{x\left(x-6\right)\left(x+6\right)}\). \(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x\left(x+6\right)}{2\left(x-3\right)}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{6}{x-6}\)+\(\dfrac{x}{6-x}\)
= \(\dfrac{6}{x-6}\)- \(\dfrac{x}{x-6}\)
= \(\dfrac{6-x}{x-6}\)
= \(\dfrac{-\left(x-6\right)}{x-6}\)
= -1
\(\dfrac{36}{x}+\dfrac{36}{x-12}=\dfrac{9}{2}\)
\(\Rightarrow72\left(x-12\right)+72x\left(x-12\right)-x\left(x-12\right)=0\)
\(\Leftrightarrow\left(x-12\right)\left(72+72x-x\right)=0\)
\(\Leftrightarrow\left(x-12\right)\left(72+71x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\72+71x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=\dfrac{-72}{71}\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 12 hoặc x = \(\dfrac{-72}{71}\) .
a) \(\dfrac{2}{x+3}+\dfrac{1}{x}\) [ MTC: x(x+3) ]
\(=\dfrac{x.2}{x\left(x+3\right)}+\dfrac{1\left(x+3\right)}{x\left(x+3\right)}\)
\(=\dfrac{2x+x+3}{x\left(x+3\right)}\)
\(=\dfrac{3x+3}{x\left(x+3\right)}\)
\(=\dfrac{3\left(x+1\right)}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\)
\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\) \(\left[MTC:2\left(x-1\right)\left(x+1\right)\right]\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-2x.2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x^2+2x+1\right)-4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x^2-2x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)}{2\left(x+1\right)}\)
a) Ta có :
\(\dfrac{2}{x+3}+\dfrac{1}{x}=\dfrac{2x+x+3}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)-2x.2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{-3x+1}{2\left(x-1\right)\left(x+1\right)}\)
c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)
\(=\dfrac{y^2-12+36}{6y\left(y-6\right)}=\dfrac{y^2-24}{6y\left(y-6\right)}\)
d) \(\dfrac{6+x}{x+3x}+\dfrac{3}{2x+6}=\dfrac{6+x}{4x}+\dfrac{3}{2\left(x+3\right)}\)
\(=\dfrac{\left(6+x\right)\left(2x+6\right)+12x}{8x\left(x+3\right)}\)(Đề câu này phải sửa thành\(\dfrac{6+x}{x^2+3x}chứ\)) ???
\(a,\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2.\left(x-1\right)}+\dfrac{-2x}{\left(x+1\right).\left(x-1\right)}=\dfrac{\left(x+1\right).\left(x+1\right)}{2.\left(x-1\right).\left(x+1\right)}+\dfrac{\left(-2x\right).x}{x.\left(x+1\right).\left(x-1\right)}=\dfrac{\left(x+1\right).\left(x+1\right)-2x^2}{x.\left(x+1\right)\left(x-1\right)}\)
b: \(=\dfrac{y^2-12y+24}{6y\left(y-6\right)}\)
c: \(=\dfrac{12-2x+3x}{2x\left(x+3\right)}=\dfrac{x+12}{2x\left(x+3\right)}\)
\(I=3\left(x^2-\dfrac{5}{3}x+1\right)\)
\(I=3\left(x^2-2.x.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2-\left(\dfrac{5}{6}\right)^2+1\right)\)
\(I=3\left[\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{36}\right]\)
\(I=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{12}\)
mình ra là \(\dfrac{11}{36}\)mà bn
bn coi lại đi
I=3x2-5x+3
I=3(x2-\(\dfrac{5}{3}\)x+1)
I=3[x2-2.x.\(\dfrac{5}{3}\)+\(\left(\dfrac{5}{6}\right)^2\)-\(\left(\dfrac{5}{6}\right)^2\)+1]
I=3(x-\(\dfrac{5}{3}\))2+\(\dfrac{11}{36}\)
I=3(x-\(\dfrac{5}{3}\))2+\(\dfrac{11}{36}\)≥\(\dfrac{11}{36}\)
vậy Min I= \(\dfrac{11}{36}\)khi x =\(\dfrac{5}{3}\)
Theo mik nghĩ là vậy á
CHÚC BN HỌC TỐT
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
a) Tớ làm luôn nhé , không chép lại đề đâu
P = \(\left[\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right].\dfrac{x\left(x+6\right)}{2x-6}\)
ĐKXĐ : x # -6 ; x # 6 ; x # 0 ; x # 3 . Khi đó , ta có :
P = \(\left[\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right]\).\(\dfrac{x\left(x+6\right)}{2x-6}\)
P = \(\dfrac{x^2-x^2+12x-36}{x-6}.\dfrac{1}{2x-6}\)
P = \(\dfrac{6\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}=\dfrac{6}{x-6}\)
b) Tương tự
b: \(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3\left(x^2+x-6\right)\)
\(\Leftrightarrow3x^2-10x+3=3x^2+3x-18\)
=>-13x=-21
hay x=21/13
c: \(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\)
=>x-100=0
hay x=100
\(\dfrac{36}{x+6}+\dfrac{36}{x-6}=4,5\)
\(\Leftrightarrow36\left(x-6\right)+36\left(x+6\right)=4,5\left(x^2-36\right)\)
\(\Leftrightarrow36x-216+36x+216=4,5x^2-162\)
\(\Leftrightarrow-4,5x^2+72x+162=0\)
\(\Leftrightarrow\left(x-18\right)\left(-4,5x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-2\end{matrix}\right.\)
bạn làm rõ hơn ở chỗ này đc ko, mk ko hiểu
⇔−4,5x2+72x+162=0⇔−4,5x2+72x+162=0
⇔(x−18)(−4,5x−9)=0