2x + 3x = 15
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\(\left(2x-5\right).\left(4-3x\right)-\left(3x+11\right).\left(5-2x\right)-15\left(2x-5\right)\)
\(=\)\(\left(2x-5\right).\left(4-3x\right)+\left(3x+11\right).\left(2x-5\right)-15\left(2x-5\right)\)
\(=\left(2x-5\right).\left(4-3x+3x+11-15\right)\)
\(=\left(2x-5\right).0\)
\(=0\)
giải
(2x-5).(4-3x)-(3x+11).(5-2x)-15.(2x-5)
8x-6x^2-20+15x-(15x-6x^2+55-22x)-30x+75
8x-6x^2-20+15x-15x+6x^2-55+22x-30x+75
(8x+15x-15x+22x-30x)+(-6x^2+6x^2)+(-20-55+75)
0+0+0=0
chúc bạn học tốt nha
ĐK: \(x\in R\)
Đặt \(\sqrt{3x^2-2x+15}=a,\sqrt{3x^2-2x+8}=b\left(a,b>0\right)\)
\(pt\Leftrightarrow a+b=a^2-b^2\)
\(\Leftrightarrow\left(a+b\right)\left(a-b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\left(l\right)\\a=b+1\end{matrix}\right.\)
\(a=b+1\)
\(\Leftrightarrow\sqrt{3x^2-2x+15}=\sqrt{3x^2-2x+8}+1\)
\(\Leftrightarrow3x^2-2x+15=3x^2-2x+8+1+2\sqrt{3x^2-2x+8}\)
\(\Leftrightarrow\sqrt{3x^2-2x+8}=3\)
\(\Leftrightarrow3x^2-2x+8=9\)
\(\Leftrightarrow3x^2-2x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\left(x+7\right)+3x\left(2x-1\right)-2x\left(3x+15\right)=-42\)
=>\(x+7+6x^2-3x-6x^2-30x=-42\)
=>\(-32x=-42-7=-49\)
=>\(x=\dfrac{49}{32}\)
1) (2x + 1)(3x – 2) = (5x – 8)(2x + 1)
⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0
⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0
⇔ (2x + 1).(3x – 2 – 5x + 8) = 0
⇔ (2x + 1)(6 – 2x) = 0
⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)
Vậy.....
2) 4x2 -1 = (2x + 1)(3x - 5)
⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0
⇔ (2x+1)(2x-1-3x+5)=0
⇔ (2x+1)(4-x)=0
⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)
Vậy...
3)
(x + 1)2 = 4(x2 – 2x + 1)
⇔ (x + 1)2 - 4(x2 – 2x + 1) = 0
⇔ x2 + 2x +1- 4x2 + 8x – 4 = 0
⇔ - 3x2 + 10x – 3 = 0
⇔ (- 3x2 + 9x) + (x – 3) = 0
⇔ -3x (x – 3)+ ( x- 3) = 0
⇔ ( x- 3) ( - 3x + 1) = 0
⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy......
a) Ta có: \(3x\left(7x-2\right)-14x+4=0\)
\(\Leftrightarrow3x\left(7x-2\right)-2\left(7x-2\right)=0\)
\(\Leftrightarrow\left(7x-2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-2=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=2\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{7};\dfrac{2}{3}\right\}\)
b) ĐKXĐ: \(x\notin\left\{0;3\right\}\)
Ta có: \(\dfrac{2x+1}{x-3}+\dfrac{5-3x}{x}=\dfrac{2x^2-15}{x^2-3x}\)
\(\Leftrightarrow\dfrac{x\left(2x+1\right)}{x\left(x-3\right)}+\dfrac{\left(5-3x\right)\left(x-3\right)}{x\left(x-3\right)}=\dfrac{2x^2-15}{x\left(x-3\right)}\)
Suy ra: \(2x^2+x+5x-15-3x^2+9x-2x^2+15=0\)
\(\Leftrightarrow-3x^2+15x=0\)
\(\Leftrightarrow-3x\left(x-5\right)=0\)
mà -3<0
nên x(x-5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\left(nhận\right)\end{matrix}\right.\)
Vậy: S={5}
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2x+2}{3x-6}=\frac{2x-6}{3x-15}=\frac{\left(2x+2\right)-\left(2x-6\right)}{\left(3x-6\right)-\left(3x-15\right)}=\frac{2x+2-2x+6}{3x-6-3x+15}=\frac{8}{9}\)
=> (2x + 2).9 = (3x - 6).8
=> 18x + 18 = 24x - 48
=> 18 + 48 = 24x - 18x
=> 6x = 66
=> x = 66 : 6 = 11
2x+3x -15=200
5x=200+15
5x=215
x=215:5
x=43
1x+2x+3x+10=500
6x=500-10
6x=490
x=490/6=245/3
/ là dấu gạch phấn số nha
2x+3x-15=200
=> 5x = 200+15
=> 5x = 215
=> x= 43
1x+2x+3x+10=500
=> 6x = 500 - 10
=> 6x = 490
=> x = 81,6
`2x + 3x = 15`
`x(2+3) =15`
`5x=15`
`x=15/5 =3`
`2x+3x=15`
`(2+3)x=15`
`5x=15`
`x=15:5=3`