Tìm x biết 8x(x-3)-8(x-1)(x+1)=20
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Tìm x, biết:
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)
\(\frac{37-x}{x+13}=\frac{5}{3}\)
\(\Rightarrow3\left(37-x\right)=5\left(x+13\right)\)
\(111-3x=5x+65\)
\(8x=46\)
\(x=\frac{23}{4}\)
Vậy..................
Đây nhé Yến ^^!:
\(\frac{x-2}{x-6}\)nhận giá trị dương
\(\Leftrightarrow\orbr{\begin{cases}..\\,,\end{cases}}\)
Ghép vào chỗ \(..\): \(\hept{\begin{cases}x-2>0\\x-6>0\end{cases}\Rightarrow\hept{\begin{cases}x>2\\x>6\end{cases}\Rightarrow}x>6}\)
Ghép vào chỗ \(,,\): \(\hept{\begin{cases}x-2< 0\\x-6< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 2\\x< 6\end{cases}\Rightarrow}x< 2}\)
Vậy x > 6 hoặc x < 2 .
Tìm x, biết:
a) x + 99:3 = 55
b) (x - 25): 15=20
c) (3.x - 15).7 = 42
d) (8x - 16)(x-5)=0
e) x.(x+1)=2+4+6+8+10+...+2500
a, \(x\) + 99: 3 = 55
\(x\) + 33 = 55
\(x\) = 55 - 33
\(x\) = 22
b, (\(x\) - 25):15 = 20
\(x\) - 25 = 20 x 15
\(x\) - 25 = 300
\(x\) = 300 + 25
\(x\) = 325
c, (3\(x\) - 15).7 = 42
3\(x\) - 15 = 42:7
3\(x\) - 15 = 6
3\(x\) = 6 + 15
3\(x\) = 21
\(x\) = 21: 3
\(x\) = 7
d, (8\(x\) - 16).(\(x\) -5) = 0
\(\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}8x=16\\x=5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=16:8\\x=5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy \(x\) \(\in\) {2; 5}
tìm x
8x(x-3)-8(x-1)(x+1)=20
cảm ơn ạ! =))
8x(x - 3) - 8(x - 1)(x + 1) = 20
=> 8x2 - 24x - 8(x2 - 1) - 20 = 0
=> 8x2 - 24x - 8x2 + 8 - 20 = 0
=> -24x = -12
=> x = 1/2
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8-20=0\Leftrightarrow-24x-12=0\Leftrightarrow-24x=12\Leftrightarrow x=\frac{12}{-24}=\frac{-1}{2}\)
Ngọc Vĩ:Bạn sai rồi.-24x=12 mới đúng nhé!
Tìm x:
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)
Ta có: 8x.(x - 3) - 8.(x - 1)(x + 1) = 20
=> 8x2 - 24x - 8.(x2 - 1) = 20
=> 8x2 - 24x - 8x2 + 8 - 20 = 0
=> -24x - 12 = 0
=> -24x = 12
=> x = -1/2
1/2 là kết quả đúng nhất
Tìm x:
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)
Áp dụng hằng đẳng thức : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(pt\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8=20\)
\(\Leftrightarrow-24x+8=20\Leftrightarrow-24x=12\Leftrightarrow x=\frac{12}{-24}=-\frac{1}{2}\)
Vậy x=-1/2
Tìm x biết:
a) x 6 + 2 x 3 +1 = 0; b) x(x - 5) = 4x - 20;
c) x 4 -2 x 2 =8-4 x 2 ; d) ( x 3 - x 2 ) - 4 x 2 + 8x-4 = 0.
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
Tìm x, biết:
a) 7x(x + 1) - 3(x + 1) =0
b) 3 ( x + 8) - x^2 - 8x = 0
c) x^2 - 10x = -25
d) x^2 - 10x = -25
a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 =>
c)
bài 1: tìm x, biết:
\(\sqrt{8x}-\sqrt{200x}+5\sqrt{x}=-20\)
\(3\sqrt{5x}-\sqrt{75x}+4\sqrt{x}=10\)
Lời giải:
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+5\sqrt{x}=-20$
$\Leftrightarrow 5\sqrt{x}-8\sqrt{2x}=-20$
$\Leftrightarrow \sqrt{x}(5-8\sqrt{2})=-20$
$\Leftrightarrow \sqrt{x}=\frac{20}{8\sqrt{2}-5}$
$\Rightarrow x=(\frac{20}{8\sqrt{2}-5})^2$
b. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{5x}-5\sqrt{3x}+4\sqrt{x}=10$
$\Leftrightarrow \sqrt{x}(3\sqrt{5}-5\sqrt{3}+4)=10$
$\Leftrightarrow \sqrt{x}=\frac{10}{3\sqrt{5}-5\sqrt{3}+4}$
$\Rightarrow x=(\frac{10}{3\sqrt{5}-5\sqrt{3}+4})^2$
Tìm x, biết:
2/(x-1)(x-3)+5/(x-3)(x-8)+12/(x-8)(x-20)-1/x-20=-3/4