ai giúp mik vs 4x(9x-81)=0
giúp mik vs
\[ 5\sqrt{\frac{9x - 27}{25}} - 7\sqrt{\frac{4x - 12}{9}} - 7\sqrt{x^2 - 9} + 18\sqrt{\frac{9x^2 - 81}{81}} = 0 \]
=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)
=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)
=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)
=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)
=>x-3=0
=>x=3
tìm x biết
4x^2-81=0
các bạn giúp mik vs mik cần gấp . Thank you các bạn
4x^2-81=0
4x^2=81
x^2=81/4
x=\(\mp\frac{9}{2}\)
Vậy............
Trả lời:
\(4x^2-81=0\)
\(\Leftrightarrow\left(2x\right)^2-9^2=0\)
\(\Leftrightarrow\left(2x-9\right)\left(2x+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-9=0\\2x+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=-\frac{9}{2}\end{cases}}}\)
Vậy x = 9/2; x = - 9/2 là nghiệm của pt.
\(\sqrt{2-x}-\sqrt{8-4x}+\sqrt{18-9x}=6\) giúp mik vs
Lời giải:
ĐKXĐ: $x\leq 2$
$\sqrt{2-x}-\sqrt{4(2-x)}+\sqrt{9(2-x)}=6$
$\Leftrightarrow \sqrt{2-x}-2\sqrt{2-x}+3\sqrt{2-x}=6$
$\Leftrightarrow (1-2+3)\sqrt{2-x}=6$
$\Leftrightarrow 2\sqrt{2-x}=6$
$\Leftrightarrow \sqrt{2-x}=3$
$\Leftrightarrow 2-x=3^2=9$
$\Leftrightarrow x=2-9=-7$ (tm)
x mũ 3 - x mũ 2 +3x - 3 > 0
x mũ 3 + x mũ 2 + 9x + 9 < 0
4x mũ 3 - 14x mũ 2+ 6x - 21<0
men nào giúp mik vs
tối nay mik phải đi hok r
mấy bạn ơi giúp mik câu này vs
3x(12x-4)-(4x-3)(9x+4)=9
3x(12x-4)-(4x-3)(9x+4) = 9
36x2 -12x-(36x2 -16x-27x-12) = 9
36x2 -12x-36x2 -16x+27x+12 = 9
-x = 9-12
-x = -3
x= -3 : -1
x= 3
vậy x= 3
\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
Giải phương trình.
giúp e với ạaa :< gấp aa :((
\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)
\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)
\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Rightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{\left(x-3\right)\left(x+3\right)}=0\Rightarrow\sqrt{x-3}-3\sqrt{\left(x-3\right)\left(x+3\right)}=0\)
\(\Rightarrow\sqrt{x-3}\left(1-3\sqrt{x+3}\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=3\sqrt{x+3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)
giải hpt(:\(3x+\sqrt{1+9x^2}\)).(y+\(\sqrt{1+y^2}\))=1
vàvà x^2+y^2-4x+8y-9=0
Giúp mik vs
Bài 1 khai triển các hằng đẳng thức
K) 4.x^2=
L) 1/9x^2 -25/16 y ^2
M)1/4x^2-4x^2
N) 4 /49 -4x^2
O) (x-3) (x+3)
P (x +4) (x -4 )
Mik cần gấp giúp mik vs
m) \(\dfrac{1}{4}x^2-4x^2=\left(\dfrac{1}{2}x-2x\right)\left(\dfrac{1}{2}x+2x\right)\)
n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)
o) \(\left(x-3\right)\left(x+3\right)=x^2-9\)
a, (4x^2 - 2)^2 = 196/81
b, (2x-4)^2016 + | x^2-4 | = 0
c, (7 - 3x)^2020 = (3x - 7)^2018
d, (4x - 7)^2 -5|7 - 4x| = 0
Giúp mik vs!
a) \(\left(4x^2-2\right)^2=\frac{196}{81}\)
<=> \(2^2\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(4\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(\left(2x^2-1\right)^2=\frac{196}{81}:4\)
<=> \(\left(2x^2-1\right)^2=\frac{49}{81}\)
<=> \(2x^2-1=\pm\sqrt{\frac{49}{81}}\)
<=> \(2x^2-1=\pm\frac{7}{9}\)
<=> \(\orbr{\begin{cases}2x^2-1=\frac{7}{9}\\2x^2-1=-\frac{7}{9}\end{cases}}\)<=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)