Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
90/x - 36/(x - 6) = 2
=>90/x - 36/(x - 6) - 2=0
=>90(x-6)/x(x-6)-36x/x(x-6)-2x(x-6)/x(x-6)=0
=>[90(x-6)-36x-2x(x-6)] / [x(x-6)]=0
=>(90x-540-36x-2x^2+12x)/ [x(x-6)]=0
=>(-2x^2+66x-540)/ [x(x-6)]=0 (*) xác định
Điều kiện để (*) xác định : x(x-6)≠0=>x≠0 hoặc x-6≠0=>x≠0 hoặc x≠6
(*)=>-2x^2+66x-540=0=>x^2-33x+270=0
=>(x^2-15x)-(18x-270)=0
=>x(x-15)-18(x-15)=0
=>(x-18)(x-15)=0
=>(x-18)=0 hoặc (x-15)=0
=>x=18 hoặc x=15 (thỏa mãn điều kiện)
Vậy pt có nghiệm x=18 hoặc x=15
\(\frac{90}{x}-\frac{36}{x-6}=2\) ĐKXĐ : \(x\ne0;x\ne6\)
\(\Leftrightarrow\frac{90\left(x-6\right)}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}=\frac{2x\left(x-6\right)}{x\left(x-6\right)}\)
\(\Leftrightarrow90x-540-36x=2x^2-12x\)
\(\Leftrightarrow66x-2x^2-540=0\)
.....................
\(\frac{90}{x}-\frac{36}{x-6}=2\) MTC = x (x-6) ĐK\(\hept{\begin{cases}x\ne0\\x\ne6\end{cases}}\)
\(\frac{90\left(x-6\right)}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}=\frac{2x\left(x-6\right)}{x\left(x-6\right)}\)
\(\frac{90x-540}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}-\frac{2x^2-12x}{x\left(x-6\right)}=0\)
\(90x-540-36x-2x^2+12x=0\)
\(-2x^2+66x-540=0\)
\(-2x^2+36x+30x-540=0\)
\(-2x\left(x-18\right)+30\left(x-18\right)=0\)
\(\left(x-18\right)\left(-2x+30\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-18=0\\-2x+30=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=18\\x=15\end{cases}}\)
vậy.....
ĐKXĐ: \(x\ne0;\) \(x\ne6\)
\(\frac{90}{x}-\frac{36}{x-6}=2\)
\(\Leftrightarrow\)\(\frac{90\left(x-6\right)}{x\left(x-6\right)}-\frac{36x}{x\left(x-6\right)}=2\)
\(\Leftrightarrow\)\(\frac{90x-540-36x}{x\left(x-6\right)}=2\)
\(\Leftrightarrow\)\(\frac{54x-540}{x\left(x-6\right)}=2\)
\(\Leftrightarrow\)\(54x-540=2x\left(x-6\right)\)
\(\Leftrightarrow\)\(27x-270=x\left(x-6\right)\)
mk lm đc có vậy thôi. tham khảo nha
\(\dfrac{90}{x}-\dfrac{36}{x-6}=2\) ( 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........
\(\frac{X-90}{10}+\frac{X-76}{12}+\frac{X-58}{14}+\frac{X-36}{16}+\frac{X-15}{17}=15\)
\(\frac{X-90}{10}+\frac{X-76}{12}+\frac{X-58}{14}+\frac{X-36}{16}+\frac{X-15}{17}-15=0\)
\(\frac{X-90}{10}-1+\frac{X-76}{12}-2+\frac{X-58}{14}-3+\frac{X-36}{16}-4+\frac{X-15}{17}-5=0\)
\(\frac{X-100}{10}+\frac{X-100}{12}+\frac{X-100}{14}+\frac{X-100}{16}+\frac{X-100}{17}=0\)
\(\left(X-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
=> X-100=0
=> X=100
vậy x=100
a) 90/x - 36/x-6 = 2.
ĐKXĐ: x≠0, x≠6.
<=> 90(x-6)-36x-2x(x-6)=0
<=> 90x-540-36x-2x2+12x=0
<=> -2x2+66x-540=0
<=> \(\left[{}\begin{matrix}x=18\left(tm\right)\\x=15\left(tm\right)\end{matrix}\right.\)
b) 1/x+1/x+10=1/12
ĐKXĐ: x≠0
<=> 1/x+1/x+10-1/12=0
<=> 12+12+10.12x-x=0
<=> 12+12+120x-x=0
<=> 119x+24=0
<=> 119x=-24
<=> x= -24/119(tm)
a)ĐKXĐ:x\(\ne\)0 x\(\ne\)6
=>90(x-6)-36x=2x(x-6)
<=>90x-540-36x=2x2-12x
<=>2x2-12x=54x-540
<=>2x2-66x+540=0
<=>x2-33x+270=0
<=>(x2-15x)-(18x-270)=0
<=>(x-15)(x-18)=0
<=>x=15(tm) hoặc x=18(tm)
b)ĐKXĐ:x\(\ne\)0 x\(\ne\)3
sai đề
c)ĐKXĐ:x\(\ne\)-2 x\(\ne\)2
=>3(x-2)-2(x+2)+8=0
<=>3x-6-2x-4+8=0
<=>x-2=0
<=>x=2(L)
Vậy PT vô nghiệm
d)ĐKXĐ: x\(\ne\)-7
=>10+8=\(\dfrac{3}{2}\)(câu này hình như đề cũng sai)
Đề bài :
\(\left(x-2\right)\left(x-4\right)\left(x+6\right)\left(x+8\right)=-36\)
\(x=+_-\sqrt{34}-2,\)
\(x=-3\sqrt{2}-2,\)
\(x=3\sqrt{2}-2\)
a) \(x^2-36=0\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)
b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)
\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)
\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)
<=> [90(x-6) - 36] tất cả chia x(x-6) = 2
<=> (54x - 540) chia x(x-6) = 2
<=> (54x - 540) = 2x(x-6)
Nhân hết ra 2 vế rồi chuyển dấu => x = 18 hoăjc x= 15