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a, (3x+1)(7x+3)=(5x-7)(3x+1)
<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0
<=> (3x+1)(7x+3-5x+7)=0
<=> (3x+1)(2x+10)=0
<=> 2(3x+1)(x+5)=0
=> 3x+1=0 hoặc x+5=0
=> x= -1/3 hoặc x=-5
Vậy...
a) (3x - 2)(4x + 5) = 0
⇔ 3x - 2 = 0 hoặc 4x + 5 = 0
1) 3x - 2 = 0 ⇔ 3x = 2 ⇔ x = 2/3
2) 4x + 5 = 0 ⇔ 4x = -5 ⇔ x = -5/4
Vậy phương trình có tập nghiệm S = {2/3;−5/4}
b) (2,3x - 6,9)(0,1x + 2) = 0
⇔ 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
1) 2,3x - 6,9 = 0 ⇔ 2,3x = 6,9 ⇔ x = 3
2) 0,1x + 2 = 0 ⇔ 0,1x = -2 ⇔ x = -20.
Vậy phương trình có tập hợp nghiệm S = {3;-20}
c) (4x + 2)(x2 + 1) = 0 ⇔ 4x + 2 = 0 hoặc x2 + 1 = 0
1) 4x + 2 = 0 ⇔ 4x = -2 ⇔ x = −1/2
2) x2 + 1 = 0 ⇔ x2 = -1 (vô lí vì x2 ≥ 0)
Vậy phương trình có tập hợp nghiệm S = {−1/2}
d) (2x + 7)(x - 5)(5x + 1) = 0
⇔ 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
1) 2x + 7 = 0 ⇔ 2x = -7 ⇔ x = −7/2
2) x - 5 = 0 ⇔ x = 5
3) 5x + 1 = 0 ⇔ 5x = -1 ⇔ x = −1/5
Vậy phương trình có tập nghiệm là S = {−7/2;5;−1/5}
1/a/\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-6\end{cases}}}\)
Vậy ...................
b/ ĐKXĐ:\(x\ne2;x\ne5\)
.....\(\Rightarrow3x^2-15x-x^2+2x+3x=0\)
\(\Leftrightarrow2x^2-10x=0\)
\(\Leftrightarrow2x\left(x-5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(nhận\right)\\x=5\left(loại\right)\end{cases}}}\)
Vậy ..............
`Answer:`
`1.`
a. \(\left(x+5\right)\left(2x+1\right)-x^2+25=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-5\end{cases}}}\)
b. \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\left(ĐKXĐ:x\ne2;x\ne5\right)\)
\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\)
\(\Leftrightarrow\frac{3x\left(x-5\right)-x\left(x-2\right)+3x}{\left(x-2\right)\left(x-5\right)}=0\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)+3x=0\)
\(\Leftrightarrow3x^2-15x-x^2+2x+3x=0\)
\(\Leftrightarrow2x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\text{(Không thoả mãn)}\end{cases}}}\)
`2.`
\(ĐKXĐ:x\ne-m-2;x\ne m-2\)
Ta có: \(\frac{x+1}{x+2+m}=\frac{x+1}{x+2-m}\left(1\right)\)
a. Khi `m=-3` phương trình `(1)` sẽ trở thành: \(\frac{x+1}{x-1}=\frac{x+1}{x+5}\left(x\ne1;x\ne-5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{x-1}=\frac{1}{x+5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=x+5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\-1=5\text{(Vô nghiệm)}\end{cases}}}\)
b. Để phương trình `(1)` nhận `x=3` làm nghiệm thì
\(\Leftrightarrow\hept{\begin{cases}\frac{3+1}{3+2-m}=\frac{3+1}{3+2-m}\\3\ne-m-2\\3\ne m-2\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{5+m}=\frac{4}{5-m}\\m\ne\pm5\end{cases}}\Leftrightarrow\hept{\begin{cases}5+m=5-m\\m\ne\pm5\end{cases}}\Leftrightarrow m=0\)
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
áp dụng các hằng đẳng thức thôi mà :)
a)\(x^2-2x+1=25\)
=>\(\left(x-1\right)^2=25\)
=>\(\orbr{\begin{cases}x-1=-5\\x-1=5\end{cases}}\)
b)\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
=>\(3\left[\left(x-1\right)^2-x\left(x-5\right)\right]=1\)
=>\(3\left(x^2-2x+1-x^2+5x\right)=1\)
=>\(3\left(3x+1\right)=1\)
=>\(3x+1=\frac{1}{3}\)
=>\(3x=\frac{-2}{3}\)
=>\(x=\frac{-2}{9}\)
c)\(\left(5-2x\right)^2-16=0\)
=>\(\left(5-2x\right)^2-4^2=0\)
=>\(\left(5-2x-4\right)\left(5-2x+4\right)=0\)
=>\(\orbr{\begin{cases}5-2x-4=0\\5-2x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}}\)
a) Ta có: \(\left(x-1\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot3\cdot\left(x-2\right)=0\)
Vì 3≠0
nên \(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: x∈{1;2}
b) Ta có: \(\left(2x+5\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\1-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-5\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{2}\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-5}{2};\frac{1}{3}\right\}\)
c) Ta có: \(\left(x+1\right)\left(2x-3\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=3\\3x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{3}{2};\frac{5}{3}\right\}\)
d) Ta có: \(6\left(x-2\right)\left(x-4\right)\left(1-7x\right)=0\)
Vì 6≠0
nên \(\left[{}\begin{matrix}x-2=0\\x-4=0\\1-7x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\\7x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\\x=\frac{1}{7}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;4;\frac{1}{7}\right\}\)
e) Ta có: \(\left(x+1\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
Vậy: x∈{-1;-2}
f) Ta có: \(\left(3x-2\right)^2\cdot\left(x+1\right)\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3x-2\right)^2=0\\x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x=-1\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-1\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{2}{3};-1;2\right\}\)
g) Ta có: \(\left(5-x\right)^2\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(5-x\right)^2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5-x=0\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{1}{3}\right\}\)
h) Ta có: \(\left(14-2x\right)^2\cdot\left(3-x\right)\cdot\left(2x-4\right)=0\)
\(\Leftrightarrow4\left(7-x\right)^2\cdot\left(3-x\right)\cdot2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow8\cdot\left(7-x\right)^2\cdot\left(3-x\right)\cdot\left(x-2\right)=0\)
Vì 8≠0
nên \(\left[{}\begin{matrix}\left(7-x\right)^2=0\\3-x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7-x=0\\x=3\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\\x=2\end{matrix}\right.\)
Vậy: x∈{7;3;2}
i) Ta có: \(\left(5x-6\right)^2\cdot\left(x+2\right)\cdot\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(5x-6\right)^2=0\\x+2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x-6=0\\x=-2\\x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-2\\x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{5}\\x=-2\\x=-10\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{6}{5};-2;-10\right\}\)
j) Ta có: \(\left(3x-3\right)^3\cdot\left(x+4\right)=0\)
\(\Leftrightarrow27\cdot\left(x-1\right)^3\cdot\left(x+4\right)=0\)
Vì 27≠0
nên \(\left[{}\begin{matrix}\left(x-1\right)^3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Vậy: x∈{1;-4}
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
