Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{2x-9}{2x-5}+\dfrac{3x}{3x-2}=2\)
\(\Rightarrow\dfrac{2x-5-4}{2x-5}+\dfrac{3x-2+2}{3x-2}=2\)
\(\Rightarrow1-\dfrac{4}{2x-5}+1+\dfrac{2}{3x-2}=2\)
\(\Rightarrow\dfrac{4}{2x-5}+\dfrac{2}{3x-2}=0\)
\(\Rightarrow\dfrac{12x-8}{\left(2x-5\right)\left(3x-2\right)}+\dfrac{4x-10}{\left(2x-5\right)\left(3x-2\right)}=0\)
\(\Rightarrow\dfrac{16x-18}{\left(2x-5\right)\left(3x-2\right)}=0\)
\(\Rightarrow16x-18=0\)
\(\Rightarrow x=\dfrac{18}{16}=\dfrac{9}{8}\)
b) \(\dfrac{x+2}{2002}+\dfrac{x+5}{1999}+\dfrac{x+201}{1803}=-3\)
\(\Rightarrow\dfrac{x+2}{2002}+1+\dfrac{x+5}{1999}+1+\dfrac{x+201}{1803}+1=0\)
\(\Rightarrow\dfrac{x+2004}{2002}+\dfrac{x+2004}{1999}+\dfrac{x+2004}{1809}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2002}+\dfrac{1}{1999}+\dfrac{1}{1809}\right)=0\)
Vì \(\left(\dfrac{1}{2002}+\dfrac{1}{1999}+\dfrac{1}{1809}\right)\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
c) \(\left(2x^2+3x+1\right)\left(2x^2+5x+3\right)=18\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)\left(x+1\right)\left(2x+3\right)=18\)
\(\Rightarrow\left(x+1\right)^2\left(2x+1\right)=18\)
Thôi, xử tiếp đi nhé :)))))
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
\(I=3\left(x^2-\dfrac{5}{3}x+1\right)\)
\(I=3\left(x^2-2.x.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2-\left(\dfrac{5}{6}\right)^2+1\right)\)
\(I=3\left[\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{36}\right]\)
\(I=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{12}\)
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
1.
|x-9|=2x+5
x<9; x-9=-2x-5
3x=4=>x=4/3(n)
x≥9; x-9=2x+5=> x=-14(l)
2.a
A=2x-5≥0<=>2x≥5; x≥5/2
1. a) / x - 9 / = 2x + 5
Do : / x - 9 / ≥ 0 ∀x
⇒2x + 5 ≥ 0
⇔ x ≥ \(\dfrac{-5}{2}\)
Bình phương cả hai vế của phương trình , ta được :
( x - 9)2 = ( 2x + 5)2
⇔ ( x - 9)2 - ( 2x + 5)2 = 0
⇔ ( x - 9 - 2x - 5)( x - 9 + 2x + 5) = 0
⇔ ( - x - 14)( 3x - 4) = 0
⇔ x = - 14 ( KTM) hoặc : x = \(\dfrac{4}{3}\) ( TM)
KL....
b) Mạn phép làm luôn , ko chép lại đề :
\(\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-5}{\left(x-3\right)\left(x+3\right)}\) ( x # 3 ; x # - 3)
⇔ 5x + 15 + 4x - 12 = x - 5
⇔ 9x + 3 = x - 5
⇔ 8x = - 8
⇔ x = -1 ( TM)
KL....
2: \(\Leftrightarrow\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)=2\left(x-1\right)\left(x+1\right)\)
=>x^2-3x-4+x^2+3x-4=2x^2-2
=>2x^2-8=2x^2-2(loại)
3: \(\Leftrightarrow\left(x^2-x\right)\left(x-3\right)+x^2\left(x+3\right)=-7x^2+3x\)
=>x^3-3x^2-x^2+3x+x^3+3x^2+7x^2-3x=0
=>2x^3+6x^2=0
=>2x^2(x+3)=0
=>x=0(nhận) hoặc x=-3(loại)
\(a.\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\\ \Leftrightarrow\dfrac{x-2}{2000}-1+\dfrac{x-3}{1999}-1=\dfrac{x-4}{1998}-1+\dfrac{x-5}{1997}-1\\ \Leftrightarrow\dfrac{x-2}{2000}-\dfrac{2000}{2000}+\dfrac{x-3}{1999}-\dfrac{1999}{1999}=\dfrac{x-4}{1998}-\dfrac{1998}{1998}+\dfrac{x-5}{1997}-\dfrac{1997}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}=\dfrac{x-2002}{1998}+\dfrac{x-2002}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}-\dfrac{x-2002}{1998}-\dfrac{x-2002}{1997}=0\\ \Leftrightarrow\left(x-2002\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \)
\(Do:\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\\ \Rightarrow x-2002=0\\ \Leftrightarrow x=2002\\ Vậy:S=\left\{2002\right\}\)
Mấy câu khác tương tự :v
b: \(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)
=>123-x=0
=>x=123
c: \(\Leftrightarrow\dfrac{x-2}{2017}+1=\dfrac{x-1}{2018}+\dfrac{x}{2019}\)
\(\Leftrightarrow\left(\dfrac{x-2}{2017}-1\right)=\left(\dfrac{x-1}{2018}-1\right)+\left(\dfrac{x}{2019}-1\right)\)
=>x-2019=0
=>x=2019
Câu 1:
Tại \(x=5\) thì ta có pt:
\(pt\Leftrightarrow10+4m^2=19\)
\(\Leftrightarrow4m^2=9\Leftrightarrow m^2=\dfrac{9}{4}\)
\(\Leftrightarrow m=\pm\sqrt{\dfrac{9}{4}}=\pm\dfrac{3}{2}\)
Vậy với \(m=\pm\dfrac{3}{2}\) thì pt có nghiệm là \(x=5\)
Câu 2:
\(\dfrac{x+5}{1999}+\dfrac{x+7}{1997}=\dfrac{x+9}{1995}+\dfrac{x+11}{1993}\)
\(\Leftrightarrow\dfrac{x+5}{1999}+1+\dfrac{x+7}{1997}+1=\dfrac{x+9}{1995}+1+\dfrac{x+11}{1993}+1\)
\(\Leftrightarrow\dfrac{x+2004}{1999}+\dfrac{x+2004}{1997}=\dfrac{x+2004}{1995}+\dfrac{x+2004}{1993}\)
\(\Leftrightarrow\dfrac{x+2004}{1999}+\dfrac{x+2004}{1997}-\dfrac{x+2004}{1995}-\dfrac{x+2004}{1993}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{1999}+\dfrac{1}{1997}-\dfrac{1}{1995}-\dfrac{1}{1993}\right)=0\)
\(\Rightarrow x+2004=0\). Do \(\dfrac{1}{1999}+\dfrac{1}{1997}-\dfrac{1}{1995}-\dfrac{1}{1993}\ne0\)
\(\Rightarrow x=-2014\)