Tìm x biết:
A.(x+3)4+(x+5)4=16
B. (x-2)4+(x-3)4=1
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a) \(x-\dfrac{5}{6}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{5}{6}\)
\(x=\dfrac{4}{3}\)
vậy x = ....
b) \(3\dfrac{1}{3}x+16\dfrac{3}{4}=-13,25\)
\(\dfrac{10}{3}\)\(x+\dfrac{67}{4}=-13,25\)
\(\dfrac{10}{3}x=\left(-13,25\right)-\dfrac{67}{4}\)
\(\dfrac{10}{3}x=-30\)
\(x=\left(-30\right):\dfrac{10}{3}\)
\(x=-9\)
vậy x =...
sai mog bn thông cảm!!!
a) x × 5 = 35 – 5
x × 5 = 30
x = 30 : 5
x = 6
b) x : 4 = 12 – 8
x : 4 = 4
x = 4 × 4
x = 16
c) 4 × x = 6 × 2
4 × x = 12
x = 12 : 4
x = 3
d) x : 3 = 16 : 4
x : 3 = 4
x = 4 × 3
x = 12
a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
a: \(\Leftrightarrow x^2-7x^2+28x=16\)
\(\Leftrightarrow-6x^2+28x-16=0\)
\(\Leftrightarrow3x^2-14x+8=0\)
\(\text{Δ}=\left(-14\right)^2-4\cdot3\cdot8=100\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-10}{6}=\dfrac{4}{6}=\dfrac{2}{3}\\x_2=\dfrac{14+10}{6}=\dfrac{24}{6}=4\end{matrix}\right.\)
a. x mũ 2 - 2x + 1 = 25
= x^2 + 2.x.1 + 1^2
= ( x + 1 ) ^2
ko bt có đúng ko nữa, mấy câu kia tui ko bt lm
a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà -3<x<30
nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)
b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)
mà -16<=x<20
nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)
c: \(\Leftrightarrow x-1+4⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
d: \(\Leftrightarrow2x+4-5⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-1;-3;3;-7\right\}\)
a) ( x + 3)4 + ( x + 5)4 = 16
Đặt : x + 4 = a , ta có :
( a - 1)4 + ( a + 1)4 = 16
=> a4 - 4a3 + 6a2 - 4a + 1 + a4 + 4a3 + 6a2 + 4a + 1 = 16
=> 2a4 + 12a2 + 2 - 16 = 0
=> 2( a4 + 6a2 - 7 ) = 0
=> a4 - a2 + 7a2 - 7 = 0
=> a2( a2 - 1) + 7( a2 - 1) = 0
=> ( a2 + 7)( a2 - 1) = 0
Do : a2 + 7 > 0 ∀a
=> a = 1 hoặc a = -1
* Với a = 1 , ta có :
x + 4 = 1
=> x = - 3
* với a = -1 , ta có :
x + 4 = -1
=> x = - 5
Vậy,...
b) ( x - 2)4 + ( x - 3)4 = 1
Đặt : x - 2 = a , ta có :
a4 + ( a - 1)4 = 1
=> a4 + a4 - 4a3 + 6a2 - 4a + 1 - 1 = 0
=> 2a4 - 4a3 + 6a2 - 4a = 0
=> 2a( a3 - 2a2 + 3a - 2) = 0
Suy ra :
*) a = 0
*) a3 - 2a2 + 3a - 2 = 0
=> a3 - a2 - a2 + a + 2a - 2 = 0
=> a2( a - 1) - a( a - 1) + 2( a - 1) = 0
=> ( a - 1)( a2 - a + 2 ) = 0
Do : a2 -a + 2 = \(a^2-2.\dfrac{1}{2}a+\dfrac{1}{4}-\dfrac{1}{4}+2=\left(a-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\text{≥}\dfrac{7}{4}>0\text{∀}a\)
=> a - 1 = 0
=> a = 1
*) Với a = 0 thì :
x - 2 = 0
=> x = 2
*) Với a = 1 , thì :
x - 2 = 1
=> x = 3
Vậy,...