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Cai nay thi co gi kho :
O trong ngoac (98734-827498+728764+123456-123456+983758734-98758700-34)=0
Vi 0 mu may , va nhan may cung = 0 nen phep tinh tren se bang 0.
Da hua la phai du loi , cho to 3 cai nha
0000000000000000000000000000000000000000000000000000000000000000000000000000
Không có giá trị nào của \(x\) và làm cho phương trình đúng.
Không có đáp án.
\(\dfrac{10^{10^{100}}}{10^{10^{100}}}=1\)
ví dụ:
\(\dfrac{10}{10}=1\)
vậy
\(x^{x^x}=10^{10^{100}}\)
a) 1,25 x 16,84 = 21,05 ; 1,2 x 15,8 = 18,96.
b) 12,48 : 0,5 x 6,25 x 4 x 2 : 2 x 3,12x 1,25 : 0,25 x 10 = 32448.
k mình nha
a: \(2x+5⋮x+1\)
=>\(2x+2+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
b: \(5x+9⋮x+2\)
=>\(5x+10-1⋮x+2\)
=>\(-1⋮x+2\)
=>\(x+2\in\left\{1;-1\right\}\)
=>\(x\in\left\{-1;-3\right\}\)
c: \(2x+11⋮x+3\)
=>\(2x+6+5⋮x+3\)
=>\(5⋮x+3\)
=>\(x+3\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-2;-4;2;-8\right\}\)
d: \(4x+9⋮2x+1\)
=>\(4x+2+7⋮2x+1\)
=>\(7⋮2x+1\)
=>\(2x+1\in\left\{1;-1;7;-7\right\}\)
=>\(2x\in\left\{0;-2;6;-8\right\}\)
=>\(x\in\left\{0;-1;3;-4\right\}\)
e: \(6x+7⋮3x+1\)
=>\(6x+2+5⋮3x+1\)
=>\(5⋮3x+1\)
=>\(3x+1\in\left\{1;-1;5;-5\right\}\)
=>\(3x\in\left\{0;-2;4;-6\right\}\)
=>\(x\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)
g: \(10x+13⋮5x+1\)
=>\(10x+2+11⋮5x+1\)
=>\(11⋮5x+1\)
=>\(5x+1\in\left\{1;-1;11;-11\right\}\)
=>\(5x\in\left\{0;-2;10;-12\right\}\)
=>\(x\in\left\{0;-\dfrac{2}{5};2;-\dfrac{12}{5}\right\}\)
(x - 4)2 = (x - 4)4
=> (x - 4)4 - (x - 4)2 = 0
=> (x - 4)2.[(x - 4)2 - 1] = 0
=> \(\left[\begin{array}{nghiempt}\left(x-4\right)^2=0\\\left(x-4\right)^2-1=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x-4=0\\\left(x-4\right)^2=1\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=4\\x-4\in\left\{1;-1\right\}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}x=4\\x\in\left\{5;3\right\}\end{array}\right.\)
Vậy \(x\in\left\{3;4;5\right\}\)
\(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Rightarrow\left(x-4\right)^2-\left(x-4\right)^4=0\)
\(\Rightarrow\left(x-4\right)^2.\left[1-\left(x-4\right)^2\right]=0\)
+) \(\left(x-4\right)^2=0\Rightarrow x-4=0\Rightarrow x=4\)
+) \(1-\left(x-4\right)^2=0\Rightarrow\left(x-4\right)^2=1\Rightarrow x-4=\pm1\)
+ \(x-4=1\Rightarrow x=5\)
+ \(x-4=-1\Rightarrow x=3\)
Vậy \(x\in\left\{4;5;3\right\}\)
x2 + 2x = 0
=> x(x + 2) = 0
=> \(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
(x - 2) + 3.x2 - 6x = 0
=> (x - 2) + 3x2 - 3x . 2 = 0
=> (x - 2) + 3x.(x - 2) = 0
=> (1 + 3x)(x - 2) = 0
=> \(\orbr{\begin{cases}1+3x=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)
cái ở trong ngoặc bằng 0
=> tích bàng 0
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theo cậu đây là lớp 1 à?