giải phương trình (3x^2 - 7x + 1)^2 = (x^2 - 3x + 5)^2
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22 tháng 1 2021
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
TM
11 tháng 8 2021
1/ \(2\left(x-5\right)=\left(-x-5\right)\)
\(\Leftrightarrow2x-10=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)
==========
2/ \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy: \(S=\left\{7\right\}\)
==========
3/ \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1=x-19\)
\(\Leftrightarrow0x=0\)
Vậy: \(S=\left\{x|x\text{ ∈ }R\right\}\)
===========
4/ \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2=10-15x\)
\(\Leftrightarrow14x=1\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy: \(S=\left\{\dfrac{1}{14}\right\}\)
==========
5/ \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x=7x+1\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy: \(S=\left\{-3\right\}\)
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Chúc bạn học tốt.
NT
11 tháng 8 2021
1. \(2\left(x-5\right)=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy \(S=\left\{\dfrac{5}{3}\right\}\)
2. \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1-x+19=0\)
\(\Leftrightarrow0x=0\)
Vậy \(S=\left\{x\in R\right\}\)
4. \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2-10+15x=0\)
\(\Leftrightarrow14x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy \(S=\left\{\dfrac{1}{14}\right\}\)
4. \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x-7x-1=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
24 tháng 4 2022
1.a)|−7x|=3x+16
Vì |-7x| ≥ 0 nên 3x+16 ≥ 0 ⇔ x ≥ \(\dfrac{-16}{3}\) (*)
Với đk (*), ta có: |-7x|=3x+16
\(\left[\begin{array}{} -7x=3x+16\\ -7x=-3x-16 \end{array} \right.\) ⇔ \(\left[\begin{array}{} -7x-3x=16\\ -7x+3x=-16 \end{array} \right.\)
⇔ \(\left[\begin{array}{} x=-1,6 (t/m)\\ x= 4 (t/m) \end{array} \right.\)
b) \(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\) = \(\dfrac{5x-8}{x^2-4}\)
⇔ \(\dfrac{(x-1)(x-2)}{x^2-4}\) - \(\dfrac{x(x+2)}{x^2-4}\) = \(\dfrac{5x-8}{x^2-4}\)
⇒ x2 - 2x - x + 2 - x2 - 2x = 5x - 8
⇔ -5x - 5x = -8 - 2
⇔ -10x = -10
⇔ x=1
2.7x+5 < 3x−11
⇔ 7x - 3x < -11 - 5
⇔ 4x < -16
⇔ x < -4
bạn tự biểu diễn trên trục số nha !
4 tháng 3 2022
a: 7x+35=0
=>7x=-35
=>x=-5
b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
hay x=7(loại)
4 tháng 3 2022
a, \(7x=-35\Leftrightarrow x=-5\)
b, đk : x khác 7
\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)
vậy pt vô nghiệm
2, thiếu đề
TM
0
NN
15 tháng 3 2019
\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)[x\left(x-4\right)-3\left(x-4\right)]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\)
Tương đương với 1 trong 3 biểu thức trên bằng 0.
Giải ra 3 nghiệm là \(x=\frac{1}{3};x=4;x=3\)
DC
2
27 tháng 2 2021
`(2x)/(3x^2-x+2)-(7x)/(3x^2+5x+2)=1(x ne -1,-2/3)`
Đặt `a=3x^2+2x+2(a>=5/3)`
`pt<=>(2x)/(a-3x)-(7x)/(a+3x)=1`
`=>2x(a+3x)-7x(a-3x)=a^2-9x^2`
`<=>2ax+6x^2-7ax+21x^2=a^2-9x^2`
`<=>-5ax+27x^2=a^2-9x^2`
`<=>a^2-36x^2+5ax=0`
`<=>a^2-4ax+9ax-36x^2=0`
`<=>a(a-4x)+9x(a-4x)=0`
`<=>(a-4x)(a+9x)=0`
`+)a=4x`
`=>3x^2+2x+2=4x`
`=>3x^2-2x+2=0`
`=>x^2-2/3x+2/3=0`
`=>(x-1/3)^2+5/9=0` vô lý
`+)a+9x=0`
`=>3x^2+2x+2+9x=0`
`=>3x^2+11x+2=0`
`=>x^2+11/3x+2/3=0`
`=>x=(-11+-\sqrt{97})/6`
27 tháng 2 2021
ĐKXĐ: \(x\ne-1;x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)(1)
\(\Leftrightarrow\dfrac{2}{3x-1+\dfrac{2}{x}}-\dfrac{7}{3x+5+\dfrac{2}{x}}=1\)
Đặt: \(3x+\dfrac{2}{x}=a\) (x khác 0) thì pt(1) trở thành:
\(\dfrac{2}{a-1}-\dfrac{7}{a+5}=1\)
\(\Leftrightarrow\dfrac{2\left(a+5\right)-7\left(a-1\right)}{\left(a-1\right)\left(a+5\right)}=1\)
\(\Leftrightarrow2\left(a+5\right)-7\left(a-1\right)=\left(a-1\right)\left(a+5\right)\)
\(\Leftrightarrow-5a+17=a^2+4a-5\)
\(\Leftrightarrow a^2+4a+5-5-17=0\)
\(\Leftrightarrow a^2+9a-22=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{2}{x}=2\\3x+\dfrac{2}{x}=-11\end{matrix}\right.\)
Vì \(\left\{{}\begin{matrix}3x^2+2-2x\ne0\\3x^2+11x+2\ne0\end{matrix}\right.\)
=> PT vô nghiệm
Ủa hình như sai:vvv
HP
11 tháng 1 2022
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
11 tháng 1 2022
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
<=>\(\orbr{\begin{cases}3x^2-7x+1=x^2-3x+5\\3x^2-7x+1=-x^2+3x-5\end{cases}}\)
<=> \(\orbr{\begin{cases}2x^2-4x-4=0\\4x^2-10x+6=0\end{cases}}\)
Sau đó áp dụng công thức nghiệm là ra