a: \(\Leftrightarrow1-x+3x+3=2x+3\)
=>2x+4=2x+3(vô lý)
b: \(\Leftrightarrow\left(x+2\right)^2-2x+3=x^2+10\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
=>4x+7=10
hay x=3/4
d: \(\Leftrightarrow\left(-2x+5\right)\left(3x-1\right)+3\left(x-1\right)\left(x+1\right)=\left(x+2\right)\left(1-3x\right)\)
\(\Leftrightarrow-6x^2+2x+15x-5+3\left(x^2-1\right)=\left(x+2\right)\left(1-3x\right)\)
\(\Leftrightarrow-6x^2+17x-5+3x^2-3=x-3x^2+2-6x\)
\(\Leftrightarrow-3x^2+17x-8=-3x^2-5x+2\)
=>22x=10
hay x=5/11
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
a) 1x−1−3x2x3−1=2xx2+x+11x−1−3x2x3−1=2xx2+x+1
Ta có: x3−1=(x−1)(x2+x+1)x3−1=(x−1)(x2+x+1)
=(x−1)[(x+12)2+34]=(x−1)[(x+12)2+34] cho nên x3 – 1 ≠ 0 khi x – 1 ≠ 0⇔ x ≠ 1
Vậy ĐKXĐ: x ≠ 1
Khử mẫu ta được:
x2+
a, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
⇔ \(6x^2-5x+3=2x-9x+6x^2\)
⇔ \(6x^2-5x+3-6x^2+9x-2x=0\)
⇔ \(2x+3=0\)
⇔ \(2x=-3\)
⇔ \(x=-\dfrac{3}{2}\)
b, \(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20\left(x-4\right)}{4.10}-\dfrac{4\left(3+2x\right)}{4.10}=\dfrac{5x}{5}+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20x-80}{40}-\dfrac{12+8x}{40}=\dfrac{5x+1-x}{5}\)
⇔ \(\dfrac{20x-80-12-8x}{40}=\dfrac{4x+1}{5}\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{4x+1}{5}=0\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{8\left(4x+1\right)}{40}=0\)
⇔ \(12x-92-8\left(4x+1\right)=0\)
⇔ 12x - 92 - 32x - 8 = 0
⇔ -100 - 20x = 0
⇔ 20x = -100
⇔ x = -100 : 20
⇔ x = -5
a: \(\Leftrightarrow5x-2+\left(2x-1\right)\left(1-x\right)=2-2x-2x^2-2x+6\)
\(\Leftrightarrow5x-2+2x-2x^2-1+x=-2x^2-4x+8\)
=>8x-3=-4x+8
=>-4x=11
hay x=-11/4
b: \(\Leftrightarrow\left(-2x+5\right)\left(3x-1\right)+3\left(x^2-1\right)=\left(x+2\right)\left(1-3x\right)\)
\(\Leftrightarrow-6x^2+2x+15x-5+3x^2-3=x-3x^2+2-6x\)
\(\Leftrightarrow17x-8=-5x+2\)
=>22x=10
hay x=5/11
a)
\(\dfrac{x-3}{5}+\dfrac{1-2x}{3}=6\\ < =>3x-9+5-10x=90\)
\(< =>3x-10x=90+9-5\\ < =>-7x=94\\ < =>x=-\dfrac{94}{7}\)
b)
\(\left(2x-3\right)\left(x^2+1\right)=0\\ < =>\left[{}\begin{matrix}2x-3=0\\x^2+1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x^2=-1\left(voli\right)\end{matrix}\right.\\ < =>x=\dfrac{3}{2}\)
c)
\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\left(x\ne-1;x\ne2\right)\)
suy ra: \(2\left(x-2\right)-x-1=3x-11\)
\(< =>2x-4-x-1-3x+11=0\)
\(< =>2x-x-3x=4+1-11\\ < =>-2x=-6\\ < =>x=3\left(tm\right)\)
a) \(\dfrac{x-3}{5}+\dfrac{1-2x}{3}=6\)
\(\Leftrightarrow3\left(x-3\right)+5\left(1-2x\right)=90\)
\(\Leftrightarrow-4-7x=90\)
\(\Leftrightarrow x=-\dfrac{94}{7}\)
b) \(\left(2x-3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x-3=0\) (Vì \(x^2+1>0\))
\(\Leftrightarrow x=\dfrac{3}{2}\)
c) \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\left(Đk:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\)
\(\Leftrightarrow x-5=3x-11\)
\(\Leftrightarrow x=3\)
câu c chưa đối chiếu điều kiện anh ơi
\(\dfrac{x-3}{5}+\dfrac{1-2x}{3}=6\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}+\dfrac{5\left(1-2x\right)}{15}=\dfrac{90}{15}\)
\(\Leftrightarrow\dfrac{3x-9+5-10x}{15}=\dfrac{90}{15}\)
\(\Leftrightarrow\dfrac{-7x-4}{15}-\dfrac{90}{15}=0\)
\(\Leftrightarrow\dfrac{-7x-4-90}{15}=0\)
\(\Leftrightarrow\dfrac{-7x-94}{15}=0\)
\(\Leftrightarrow-7x-94=0\)
\(\Leftrightarrow-7x=94\)
\(\Leftrightarrow x=-\dfrac{94}{7}\)
vậy phương trình có nghiệp `x=-94/7`
`-----------`
\(\left(2x-3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\x^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x^2=-1\left(loaij\right)\end{matrix}\right.\)
vậy phương trình có nghiệm `x=3/2`
`--------`
\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
ta có : \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\)
\(\Leftrightarrow2x-4-x-1=3x-11\)
\(\Leftrightarrow x-5=3x-11\)
\(\Leftrightarrow x-3x=-11+5\)
\(\Leftrightarrow-2x=-6\)
\(\Leftrightarrow x=3\)
vậy phương trình có nghiệm `x=3`
a) �−35+1+2�3=6⇔3�−9+5+10�=90⇔13�=94⇔�=94135x−3+31+2x=6⇔3x−9+5+10x=90⇔13x=94⇔x=1394.
Vậy tập nghiệm của phương trình đã cho là �={9413}S={1394}.
b) (2�−3)(�2+1)=0⇔[2�−3=0�2+1=0⇔2�−3=0⇔�=32(2x−3)(x2+1)=0⇔[2x−3=0x2+1=0⇔2x−3=0⇔x=23.
Vậy tập nghiệm của phương trình đã cho là �={32}S={23}.
c)
2�+1−1�−2=3�−11(�+1)(�−2)(đk: �≠−1,�≠2)⇒2(�−2)−(�+1)=3�−11⇔2�−4−�−1=3�−11⇔−2�=−6⇔�=3(thỏa ma˜n)⇒⇔⇔⇔x+12−
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a)
�−35+1−2�3=6<=>3�−9+5−10�=905x−3+31−2x=6<=>3x−9+5−10x=90
<=>3�−10�=90+9−5<=>−7�=94<=>�=−947<=>3x−10x=90+9−5<=>−7x=94<=>x=−794
b)
(2�−3)(�2+1)=0<=>[2�−3=0�2+1=0<=>[�=32�2=−1(����)<=>�=32(2x−3)(x2+1)=0<=>[2x−3=0x2+1=0<=>[x=23x2=−1(voli)<=>x=23
c)
2�+1−1�−2=3�−11(�+1)(�−2)(�≠−1;�≠2)x+12−x−21
a)
�−35+1−2�3=6<=>3�−9+5−10�=905x−3+31−2x=6<=>3x−9+5−10x=90
<=>3�−10�=90+9−5<=>−7�=94<=>�=−947<=>3x−10x=90+9−5<=>−7x=94<=>x=−794
b)
(2�−3)(�2+1)=0<=>[2�−3=0�2+1=0<=>[�=32�2=−1(����)<=>�=32(2x−3)(x2+1)=0<=>[2x−3=0x2+1=0<=>[x=23x2=−1(voli)<=>x=23
c)
2�+1−1�−2=3�−11(�+1)(�−2)(�≠−1;�≠2)x+12−x−21
a) �−35+1+2�3=6⇔3�−9+5+10�=90⇔13�=94⇔�=94135x−3+31+2x=6⇔3x−9+5+10x=90⇔13x=94⇔x=1394.
Vậy tập nghiệm của phương trình đã cho là �={9413}S={1394}.
b) (2�−3)(�2+1)=0⇔[2�−3=0�2+1=0⇔2�−3=0⇔�=32(2x−3)(x2+1)=0⇔[2x−3=0x2+1=0⇔2x−3=0⇔x=23.
Vậy tập nghiệm của phương trình đã cho là �={32}S={23}.
c)
2�+1−1�−2=3�−11(�+1)(�−2)(đk: �≠−1,�≠2)⇒2(�−2)−(�+1)=3�−11⇔2�−4−�−1=3�−11⇔−2�=−6⇔�=3(thỏa ma˜n)⇒⇔⇔⇔x+12−
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