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a,\(4x\left(2x+3\right)-x\left(8x-1\right)=5\left(x+2\right)\)
\(< =>8x^2+12x-8x^2+x=5x+10\)
\(< =>13x=5x+10< =>8x=10\)
\(< =>x=\frac{10}{8}=\frac{5}{4}\)
b, \(\left(3x-5\right)\left(3x+5\right)-x\left(9x-1\right)=4\)
\(< =>9x^2-25-9x^2+x=4\)
\(< =>x=4+29=33\)
c,\(3-4x\left(25-2x\right)=8x^2+x-300\)
\(< =>3-100x+8x^2=8x^2+x-300\)
\(< =>x+100x=3+300\)
\(< =>101x=303< =>x=\frac{303}{101}=3\)
d,\(2\left(1-\frac{3x}{5}\right)-\frac{2+3x}{10}=7-\frac{3\left(2x+1\right)}{4}\)
\(< =>2-\frac{6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)
\(< =>-\frac{24x}{20}-\frac{4+6x}{20}+\frac{30x+15}{20}=5\)
\(< =>\frac{30x-6x-24x+15-4}{20}=5\)
\(< =>\frac{11}{5}=5< =>11=25\)(vo li)
=) vào ngay quả bảng phá dấu GTTĐ, cay thế :<
a, \(3x+\frac{2x}{3}-3=\frac{5}{2}x-2\Leftrightarrow\frac{18x+4x-18}{6}=\frac{15x-12}{6}\)
\(\Rightarrow22x-18=15x-12\Leftrightarrow7x=6\Leftrightarrow x=\frac{6}{7}\)
Vậy pt có nghiệm x = 6/7
b, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
\(\Leftrightarrow\frac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\frac{x+7}{12}\)
\(\Rightarrow18x+9-10x-6+4x+4=x+7\)
\(\Leftrightarrow12x+7=x+7\Leftrightarrow11x=0\Leftrightarrow x=0\)
Vậy pt có nghiệm là x = 0
c, \(\frac{3x}{x-3}-\frac{x-3}{x+3}=2\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow\frac{3x\left(x+3\right)-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow3x^2+9x-x^2+6x-9=2\left(x^2-9\right)\)
\(\Leftrightarrow2x^2+15x-9=2x^2-18\Leftrightarrow15x+9=0\Leftrightarrow x=-\frac{9}{15}=-\frac{3}{5}\)
Vậy pt có nghiệm là x = -3/5
d, Sửa đề : \(\frac{x+10}{2003}+\frac{x+6}{2007}+\frac{x+2}{2011}+3=0\)
\(\Leftrightarrow\frac{x+10}{2003}+1+\frac{x+6}{2007}+1+\frac{x+2}{2011}+1=0\)
\(\Leftrightarrow\frac{x+2013}{2003}+\frac{x+2013}{2007}+\frac{x+2013}{2011}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2003}+\frac{1}{2007}+\frac{1}{2011}\ne0\right)=0\Leftrightarrow x=-2013\)
Vậy pt có nghiệm là x = -2013
e, \(4\left(x+5\right)-3\left|2x-1\right|=10\)
\(\Leftrightarrow4x+20-3\left|2x-1\right|=10\Leftrightarrow-3\left|2x-1\right|=-10-4x\)
\(\Leftrightarrow\left|2x-1\right|=\frac{10+4x}{3}\)
ĐK : \(\frac{10+4x}{3}\ge0\Leftrightarrow10+4x\ge0\Leftrightarrow x\ge-\frac{10}{4}=-\frac{5}{2}\)
TH1 : \(2x-1=\frac{10+4x}{3}\Rightarrow6x-3=10+4x\Leftrightarrow2x=13\Leftrightarrow x=\frac{13}{2}\)( tm )
TH2 : \(2x-1=\frac{-10-4x}{3}\Rightarrow6x-3=-10-4x\Leftrightarrow10x=-7\Leftrightarrow x=-\frac{7}{10}\)( tm )
f, để mình xem lại đã, quên cách phá GTTĐ rồi :v :>
Nhiều vậy ai làm hết được :P
1) \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)
\(\Leftrightarrow\frac{3x-8}{3}=\frac{4x-1}{4}\)
\(\Leftrightarrow4\left(3x-8\right)=3\left(4x-1\right)\)
\(\Leftrightarrow12x-32=12x-3\)(vô lí)
Vậy pt vô nghiệm
P/s: mấy câu sau tương tự thôi mà :)))
nhăm nhe 1 câu thôi
\(10,\frac{3+5x}{5}-3=\frac{9x-3}{4}\)
\(\Leftrightarrow\frac{3+5x-15}{5}=\frac{9x-3}{4}\)
\(\Leftrightarrow\frac{-12+5x}{5}=\frac{9x-3}{4}\)
\(\Leftrightarrow\left(-12+5x\right)5=\left(9x-3\right)4\)
\(\Leftrightarrow-60+25x=36x-12\)
\(\Leftrightarrow26x-36x=-12+60\)
\(\Leftrightarrow-10x=48\)
\(\Leftrightarrow x=-4,8\)
a) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
\(\Leftrightarrow\frac{3\left(x-3\right)}{15}=\frac{90-5\left(1-2x\right)}{15}\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow-7x=94\)
\(\Leftrightarrow x=-\frac{94}{7}\)
b) \(\frac{3x-2}{6}-5=3-\frac{2\left(x+7\right)}{4}\)
\(\Leftrightarrow\frac{2\left(3x-2\right)-60}{12}=\frac{36-6\left(x+7\right)}{12}\)
\(\Leftrightarrow6x-4-60=36-6x-42\)
\(\Leftrightarrow12x=52\)
\(\Leftrightarrow x=\frac{29}{6}\)
c) \(\frac{3x-9}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)
\(\Leftrightarrow\frac{5\left(3x-9\right)}{20}+\frac{2\left(4x-10,5\right)}{20}=\frac{12\left(x+1\right)+120}{20}\)
\(\Leftrightarrow15x-45+8x-21=12x+12+120\)
\(\Leftrightarrow23x-66=12x+132\)
\(\Leftrightarrow11x=198\)
\(\Leftrightarrow x=18\)
Chúc cậu học tốt !
a, \(-\left(x+3\right)\left(x-4\right)+\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow-\left(x^2-4x+3x-12\right)+x^2-1=10\)
\(\Rightarrow-x^2+x+12+x^2-1=10\)
\(\Rightarrow x=10+1-12\Rightarrow x=-1\)
b, \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)
\(\Rightarrow2x^2-4x-x+2-\left(2x^2-7x+6x-21\right)=3\)
\(\Rightarrow2x^2-5x+2-2x^2+x+21=3\)
\(\Rightarrow-4x=3-21-2\Rightarrow-4x=-20\)
\(\Rightarrow x=5\)
Các câu còn lại làm tương tự! Phá ngoặc ra!
Chúc bạn học tốt!!!
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)
\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)
\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)
\(< =>45x-90+50x-150+12x-60=0\)
\(< =>107x-300=0< =>x=\frac{300}{107}\)
\(\frac{x-3}{5.7}+\frac{x-5}{3.7}+\frac{x-7}{3.5}=2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\right)\)\(< =>\frac{\left(x-3\right).3}{3.5.7}+\frac{\left(x-5\right).5}{3.5.7}+\frac{\left(x-7\right).7}{3.5.7}=2\left(\frac{5.7}{3.5.7}+\frac{3.7}{3.5.7}+\frac{5.3}{3.5.7}\right)\)
\(< =>3x-9+5x-25+7x-49=2\left(35+21+15\right)\)
\(< =>15x=2.71+25+49+9\)
\(< =>15x=142+83=225< =>x=\frac{225}{15}=15\)
1
a)Ta có:
$ \dfrac{3(2x+1)}{4}-\dfrac{5x+3}{6}=\dfrac{2x-1}{3}-\dfrac{3-x}{4} $
Nhân cả hai vế với $12$:
$9(2x+1)-2(5x+3)=4(2x-1)-3(3-x)$
$18x+9-10x-6=8x-4-9+3x$
$8x+3=11x-13$
$3x=16$
$x=\dfrac{16}{3}$
Vậy $x=\dfrac{16}{3}$.
b)Ta có:
$ \dfrac{19}{4}-\dfrac{2(3x-5)}{5}=\dfrac{3-2x}{10}-\dfrac{3x-1}{4} $
Nhân cả hai vế với $20$:
$95-8(3x-5)=2(3-2x)-5(3x-1)$
$95-24x+40=6-4x-15x+5$
$135-24x=11-19x$
$124=5x$
$x=\dfrac{124}{5}$
Vậy $x=\dfrac{124}{5}$.
c)Ta có:
$ \dfrac{3(x-2)}{5}+\dfrac{5(x-3)}{8}+\dfrac{2(x-5)}{7}=10 $
Nhân cả hai vế với $280$:
$168(x-2)+175(x-3)+80(x-5)=2800$
$168x-336+175x-525+80x-400=2800$
$423x-1261=2800$
$423x=4061$
$x=\dfrac{4061}{423}$
Vậy $x=\dfrac{4061}{423}$.
d)Ta có:
$ \dfrac{x-3}{5\cdot7}+\dfrac{x-5}{3\cdot7}+\dfrac{x-7}{3\cdot5}=2\left(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}\right) $
$ \dfrac{x-3}{35}+\dfrac{x-5}{21}+\dfrac{x-7}{15}=2\left(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}\right) $
Nhân cả hai vế với $105$:
$3(x-3)+5(x-5)+7(x-7)=70+42+30$
$3x-9+5x-25+7x-49=142$
$15x-83=142$
$15x=225$
$x=15$
Vậy $x=15$.
Ta có:
$ \dfrac{x-1}{9}+\dfrac{x-2}{8}=\dfrac{x-3}{7}+\dfrac{x-4}{6} $
Nhân cả hai vế với $504$:
$56(x-1)+63(x-2)=72(x-3)+84(x-4)$
$56x-56+63x-126=72x-216+84x-336$
$119x-182=156x-552$
$370=37x$
$x=10$
Vậy $x=10$.
b)Ta có:
$ \left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdots+\dfrac{1}{9\cdot10}\right)(x-1)+\dfrac{x}{10}=x-\dfrac{9}{10} $
Ta có:
$ \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\cdots+\dfrac{1}{9\cdot10}=\dfrac{1}{1}-\dfrac{1}{10}=\dfrac{9}{10} $
Suy ra:
$ \dfrac{9}{10}(x-1)+\dfrac{x}{10}=x-\dfrac{9}{10} $
$ \dfrac{9x}{10}-\dfrac{9}{10}+\dfrac{x}{10}=x-\dfrac{9}{10} $
$x-\dfrac{9}{10}=x-\dfrac{9}{10}$
Vậy phương trình có vô số nghiệm: $x\in\mathbb{R}$.