2 Giải các phương trình
c. \(\dfrac{x-1}{5}\)+x = \(\dfrac{x+1}{7}\) d.2(x-2.5)=0.25+\(\dfrac{4x-3}{8}\)
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a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
b) Ta có: \(3+\left(x-5\right)=2\left(3x-2\right)\)
\(\Leftrightarrow3+x-5=6x-4\)
\(\Leftrightarrow x-2-6x+4=0\)
\(\Leftrightarrow-5x+2=0\)
\(\Leftrightarrow-5x=-2\)
\(\Leftrightarrow x=\dfrac{2}{5}\)
Vậy: \(S=\left\{\dfrac{2}{5}\right\}\)
c) Ta có: \(2\left(x-0.5\right)+3=0.25\left(4x-1\right)\)
\(\Leftrightarrow2x-1+3=x-\dfrac{1}{4}\)
\(\Leftrightarrow2x+2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x+\dfrac{9}{4}=0\)
\(\Leftrightarrow x=-\dfrac{9}{4}\)
Vậy: \(S=\left\{-\dfrac{9}{4}\right\}\)
d) Ta có: \(2\left(x-\dfrac{1}{4}\right)-4=-6\left(-\dfrac{1}{3}x+0.5\right)+2\)
\(\Leftrightarrow2x-\dfrac{1}{2}-4=2x-3+2\)
\(\Leftrightarrow2x-\dfrac{9}{2}=2x-1\)
\(\Leftrightarrow2x-2x=-1+\dfrac{9}{2}\)
\(\Leftrightarrow0x=\dfrac{7}{2}\)(vô lý)
Vậy: \(S=\varnothing\)
TK
https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5
a: \(\Leftrightarrow4x-5=2x-2+x\)
=>4x-5=3x-2
=>x=3(nhận)
b: =>7x-35=3x+6
=>4x=41
hay x=41/4(nhận)
c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
=>-6x+16=-5x+11
=>-x=-5
hay x=5(nhận)
d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)
\(\Leftrightarrow4x=16\)
hay x=4(nhận)
\(c,\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+10y=2\\6x-3y=-24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}13y=26\\6x-3y=-24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\6x-3.2=-24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-3\end{matrix}\right.\)
\(d,\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\left(I\right)\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\left(x\ne0\right)\\\dfrac{1}{y}=b\left(y\ne0\right)\end{matrix}\right.\)
\(\left(I\right)\Rightarrow\left\{{}\begin{matrix}a-b=1\\3a+4b=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a-3b=3\\3a+4b=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-7b=-2\\3a+4b=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{7}\\3a+4.\dfrac{2}{7}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{7}\\a=\dfrac{9}{7}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{2}{7}\Leftrightarrow x=\dfrac{7}{2}\\\dfrac{1}{y}=\dfrac{9}{7}\Leftrightarrow y=\dfrac{7}{9}\end{matrix}\right.\)
c. \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+10y=2\\6x-3y=-24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13y=26\\2x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-3\end{matrix}\right.\)
d. \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\left(x\ne0\right)\\\dfrac{1}{y}=b\left(y\ne0\right)\end{matrix}\right.\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}a-b=1\\3a+4b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-3b=3\\3a+4b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-7b=-2\\a-b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{7}\\a=\dfrac{9}{7}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{9}{7}\\\dfrac{1}{y}=\dfrac{2}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{9}\\y=\dfrac{7}{2}\end{matrix}\right.\)
a) ĐKXD: x ≠ 2
\(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{3-x}{x-2}=-3\)
\(\Leftrightarrow\dfrac{1-3+x}{x-2}=-3\)
\(\Leftrightarrow\dfrac{-2+x}{x-2}=-3\)
\(\Leftrightarrow-2+x=-3\left(x-2\right)\)
\(\Leftrightarrow-2+x=-3x+6\)
\(\Leftrightarrow x+3x=6+2\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\) (loại vì không thỏa mãn điều kiện)
Vậy S = ∅
b) ĐKXĐ: x ≠ 7
\(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
\(\Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{1}{x-7}=8\)
\(\Leftrightarrow\dfrac{7-x}{x-7}=8\)
\(\Leftrightarrow-1=8\left(vô-lý\right)\)
Vậy S = ∅
P/s: Ko chắc ạ!
c) ĐKXĐ: x ≠ 1
\(\dfrac{1}{x-1}+\dfrac{2x}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\)
Quy đồng và khử mẫu ta được:
\(x^2+x+1+2x\left(x-1\right)=3x^2\)
\(\Leftrightarrow x^2+x+1+2x^2-2x-3x^2=0\)
\(\Leftrightarrow-x+1=0\)
\(\Leftrightarrow x=1\) (loại vì ko t/m đk)
Vậy S = ∅
a) ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-2x=-2+9\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\dfrac{7}{19}\)
Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)
a, đk : x khác 5;-6
\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm)
b, đk : x khác 1;3
\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)
pt vô nghiệm
a, đk : x khác 5;-6
x2+12x+36+x2−10x+25=2x2+23x+61x2+12x+36+x2−10x+25=2x2+23x+61
⇔2x+61=23x+61⇔21x=0⇔x=0⇔2x+61=23x+61⇔21x=0⇔x=0(tm)
b, đk : x khác 1;3
x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)
pt vô nghiệm
c) \(\dfrac{x-1}{5}+x=\dfrac{x+1}{7}\)
\(\Leftrightarrow\dfrac{7x-7}{35}+\dfrac{35x}{35}=\dfrac{5x+5}{35}\)
\(\Rightarrow7x-7+35x=5x+5\)
\(\Leftrightarrow7x+35x-5x=5+7\)
\(\Leftrightarrow37x=12\)
\(\Leftrightarrow x=\dfrac{12}{37}\)
Vậy pt có nghiệm duy nhất \(x=\dfrac{12}{37}\)
d) \(2\left(x-2,5\right)=0,25+\dfrac{4x-3}{8}\)
\(\Leftrightarrow\dfrac{16\left(x-2,5\right)}{8}=\dfrac{2}{8}+\dfrac{4x-3}{8}\)
\(\Rightarrow16x-40=2+4x-3\)
\(\Leftrightarrow16x-4x=2-3+40\)
\(\Leftrightarrow12x=39\)
\(\Leftrightarrow x=3,25\)
Vậy pt có nghiệm duy nhất \(x=3,25\)