1+1= ......
0 x 9999999999999999999 = ....................................................
9999999+9999999999999999999
3090*9999999999999999999
9999999999999999999 cm2 = ......................................km2
999999999,9999999999km2 k cho minh nha
đảm bảo đúng 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000% lun
9999999999999999999999999999999999999999999999999999999999999999999999999999999 nha đúng 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000%
9999999999999999999999999999999999999999999999999999999999999 nhân 9999999999999999999 bằng ?
1+2=?
1+2+5=?
1=2+?
9999999999999999999999999999-9999=?
1+2+3+4+5+6+7+9999999999999999999=?
viết một bài văn (bao nhiêu câu cũng đc) nói về người thân đi làm xa rồi về ăn tết với gia đình
Ai trả lời được câu hỏi này mình sẽ like:9999999999999999999-53738289202763664748594987263849957748405058737850506974839203948899999999999999999999999999999999999999999999 bằng mấy
−5.3738289202763664748594987263849957748405058737850506974839203948899999999999999999999999999999999999999999999
-5.373289202763664748594987263849957748405058737850506974839203948899999999999999999999999999999999999999999999999999999999999999
|x-3| + |y-2x | =0
|x| + 3|2x -x ² | =0
|5x ² -5 | + 4|y-7 | =0
||x +1 | + |y-5 |=0
|x ² -1| + |y-1| =0
|x-1 | +|x ²-x |=0
a) Ta có: \(\left|x-3\right|+\left|y-2x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x=2\cdot3=6\end{matrix}\right.\)
bạn tin lúc trước tớ nói không tớ sai ở chổ 1x0 đóooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
|x-3| + |y-2x | =0 |x| + 3|2x -x ² | =0 |5x ² -5 | + 4|y-7 | =0 ||x +1 | + |y-5 |=0 |x ² -1| + |y-1| =0 |x-1 | +|x ²-x |=0
a) Ta có: \(\left|x-3\right|+\left|y-2x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x=2\cdot3=6\end{matrix}\right.\)
Bài 2: Tìm x
a) (x-2)2-(2x+3)2=0 d) x2.(x+1)-x.(x+1)+x.(x-1)=0
b) 9.(2x+1)2-4.(x+1)2=0 e) (x-2)2-(x-2).(x+2)=0
c) x3-6x2+9x=0 g) x4-2x2+1=0
h) 4x2+y2-20x-2y+26=0 i) x2-2x+5+y2-4y=0
a: \(\left(x-2\right)^2-\left(2x+3\right)^2=0\)
=>(x-2-2x-3)(x-2+2x+3)=0
=>(-x-5)(3x+1)=0
=>(x+5)(3x+1)=0
=>\(\left[\begin{array}{l}x+5=0\\ 3x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=-\frac13\end{array}\right.\)
b: \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)
=>\(\left\lbrack3\left(2x+1\right)\right\rbrack^2-\left\lbrack2\left(x+1\right)\right\rbrack^2=0\)
=>\(\left(6x+3\right)^2-\left(2x+2\right)^2=0\)
=>(6x+3+2x+2)(6x+3-2x-2)=0
=>(8x+5)(4x+1)=0
=>\(\left[\begin{array}{l}8x+5=0\\ 4x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac58\\ x=-\frac14\end{array}\right.\)
c: \(x^3-6x^2+9x=0\)
=>\(x\left(x^2-6x+9\right)=0\)
=>\(x\left(x-3\right)^2=0\)
=>\(\left[\begin{array}{l}x=0\\ \left(x-3\right)^2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=3\end{array}\right.\)
d: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
=>\(\left(x+1\right)\left(x^2-x\right)+x\left(x-1\right)=0\)
=>x(x+1)(x-1)+x(x-1)=0
=>x(x-1)(x+1+1)=0
=>x(x-1)(x+2)=0
=>\(\left[\begin{array}{l}x=0\\ x-1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-2\end{array}\right.\)
e: \(\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
=>(x-2)(x-2-x-2)=0
=>-4(x-2)=0
=>x-2=0
=>x=2
g: \(x^4-2x^2+1=0\)
=>\(\left(x^2-1\right)^2=0\)
=>\(x^2-1=0\)
=>\(x^2=1\)
=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)
h: \(4x^2+y^2-20x-2y+26=0\)
=>\(4x^2-20x+25+y^2-2y+1=0\)
=>\(\left(2x-5\right)^2+\left(y-1\right)^2=0\)
=>\(\begin{cases}2x-5=0\\ y-1=0\end{cases}\Rightarrow\begin{cases}x=\frac52\\ y=1\end{cases}\)
i: \(x^2-2x+5+y^2-4y=0\)
=>\(x^2-2x+1+y^2-4y+4=0\)
=>\(\left(x-1\right)^2+\left(y-2\right)^2=0\)
=>\(\begin{cases}x-1=0\\ y-2=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=2\end{cases}\)