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Ta có:\(8x^2+10x+3=\left(8x^2+6x\right)+\left(4x+3\right)\)
\(=2x\left(4x+3\right)+\left(4x+3\right)\)
\(=\left(2x+1\right)\left(4x+3\right)\)
\(4x^2+7x+3=\left(4x^2+4x\right)+\left(3x+3\right)\)
\(=4x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(4x+3\right)\)
\(ĐKXĐ:x\ne-1,x\ne\frac{-3}{4}\)
\(8x^2+10x+3=\frac{1}{4x^2+7x+3}\)
<=>\(\left(8x^2+10x+3\right)\left(4x^2+7x+3\right)=1\)
<=>\(\left(2x+1\right)\left(4x+3\right)\left(x+1\right)\left(4x+3\right)=1\)
<=>\(\left(2x+1\right)\left(4x+3\right)^2\left(x+1\right)=1\)
<=>\(\left(4x+2\right)\left(4x+3\right)^2\left(4x+4\right)=8\)
(Nhân cả 2 vế với 8)
<=>\(\left[\left(4x+2\right)\left(4x+4\right)\right]\left(4x+3\right)^2=8\)
<=>\(\left(16x^2+24x+8\right)\left(16x^2+24x+9\right)=8\)
Đặt \(16x^2+24x+8.5=y\)
\(ĐK:y>-0.5\)
(Vì \(16x^2+24x+8.5=\left(4x+3\right)^2-0.5>-0.5\)với mọi x thỏa mãn ĐKXĐ)
Phương trình trở thành:
(y-0.5)(y+0.5)=8
<=>\(y^2-0.25=8\)
<=>\(y^2=8.25\)
<=>\(\orbr{\begin{cases}y=\frac{\sqrt{33}}{2}\left(\text{thỏa mãn}\right)\\y=\frac{-\sqrt{33}}{2}\left(\text{loại}\right)\end{cases}}\)
Với \(y=\frac{\sqrt{33}}{2}\)
Ta có:\(16x^2+24x+8.5=\frac{\sqrt{33}}{2}\)
<=>\(32x^2+48x+17-\sqrt{33}=0\)
<=>\(\left(x\sqrt{33}+3\sqrt{2}\right)^2=\sqrt{33}+1\)
<=>\(\orbr{\begin{cases}x\sqrt{33}+3\sqrt{2}=\sqrt{\sqrt{33}+1}\\x\sqrt{33}+3\sqrt{2}=-\sqrt{\sqrt{33+1}}\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{\sqrt{\sqrt{33}+1}-3\sqrt{2}}{\sqrt{33}}\\x=\frac{-\sqrt{\sqrt{33}+1}-3\sqrt{2}}{\sqrt{33}}\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{\sqrt{33\sqrt{33}+33}-3\sqrt{66}}{33}\left(\text{thỏa mãn ĐKXĐ}\right)\\x=\frac{-\sqrt{33\sqrt{33}+33}-3\sqrt{66}}{33}\left(\text{thỏa mãn ĐKXĐ}\right)\end{cases}}\)
(Kết luận: Vậy tập nghiệm của phương trình đã cho là...)
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a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(-2x^2-5x+2\)
\(=-2\left(x^2+\dfrac{5}{2}x-1\right)\)
\(=-2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{41}{16}\right)\)
\(=-2\left(x+\dfrac{5}{4}\right)^2+\dfrac{41}{8}\)
b: \(4x^2-6x-1\)
\(=4\left(x^2-\dfrac{3}{2}x-\dfrac{1}{4}\right)\)
\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{13}{16}\right)\)
\(=4\left(x-\dfrac{3}{4}\right)^2-\dfrac{13}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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}
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>-x+2x=3-7
=>x=-4
b: =>6x+2+2x-5=0
=>8x-3=0
hay x=3/8
c: =>5x+2x-2-4x-7=0
=>3x-9=0
hay x=3
d: =>10x2-10x2-15x=15
=>-15x=15
hay x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
a, <=> x = -4
b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8
c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2
d, <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1
a, <=> x = -4
b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8
c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2
d <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{5;-2\right\}\)
\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)
\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)
Câu d xem lại đề
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3x-2=2x-3\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b) \(3-4y+24+6y=y+27+3y\)
\(\Leftrightarrow-2y=0\Leftrightarrow y=0\)
c) \(7-2x=22-3x\)
\(\Leftrightarrow x-15=0\)
\(\Leftrightarrow x=15\)
d) \(8x-3=5x+12\)
\(\Leftrightarrow3x-15=0\Leftrightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3x2 + 2x - 1 = 0
=> 3x2 + 3x - x - 1 = 0
=> 3x(x + 1) - (x + 1) = 0
=> (3x - 1)(x + 1) = 0
=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
x2 - 5x + 6 = 0
=> x2 - 2x - 3x + 6 = 0
=> x(x - 2) - 3(x - 2) = 0
=> (x - 3)(x - 2) = 0
=> \(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
3x2 + 7x + 2 = 0
=> 3x2 + 6x + x + 2 = 0
=> 3x(x + 2) + (x + 2) = 0
=> (3x + 1)(x + 2) = 0
=> \(\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
1, \(3x^2+2x-1=0\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)
2, \(x^2-5x+6=0\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
3, \(3x^2+7x+2=0\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}}\)