Tìm x :
b) 6 x + 1 2 = 2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
$x(x+2)(x+4)(x+6)+8$
$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$
$=a(a+8)+8$ (đặt $x^2+6x=a$)
$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$
Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$
2.
$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$
$=5-(x^2+5x-6)(x^2+5x+6)$
$=5-[(x^2+5x)^2-6^2]$
$=41-(x^2+5x)^2\leq 41$
Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$
a: Ta có: \(\left(x-5\right)\left(x+3\right)=x\left(x-3\right)\)
\(\Leftrightarrow x^2-2x-15-x^2+3x=0\)
\(\Leftrightarrow x=15\)
b: Ta có: \(\left(x+2\right)^2=\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow x+2=0\)
hay x=-2
c: Ta có: \(\left(x-6\right)\left(x+6\right)=x^2\)
\(\Leftrightarrow x^2-36=x^2\)(vô lý)
a. (x - 5)(x + 3) = x(x - 3)
<=> x2 + 3x - 5x - 15 = x2 - 3x
<=> x2 - x2 + 3x - 5x + 3x - 15 = 0
<=> x = 15
b. (x + 2)2 = (x - 1)(x + 2)
<=> x2 + 4x + 4 = x2 + 2x - x - 2
<=> x2 - x2 + 4x - 2x + x = -2 - 4
<=> 3x = -5
<=> \(x=\dfrac{-5}{3}\)
c. (x - 6)(x + 6) = x2
<=> x2 - 36 - x2 = 0
<=> x2 - x2 = 36
<=> 0 = 36 (vô lí)
Vậy nghiệm của PT là \(S=\varnothing\)
d. (2x - 3)2 = 4x2 - 8
<=> 4x2 - 12x + 9 - 4x2 + 8 = 0
<=> 4x2 - 4x2 - 12x = -8 - 9
<=> -12x = -17
<=> \(x=\dfrac{17}{12}\)
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)