( x + 4 ) . ( 23 - x ) = 0 tìm x
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
240-[23+(13+24.3-x)]=132
240-[23+(13+168-x)]=132
240-[23+(181-x)]=132
|
x-8:4-(46-23.2+6.3)=0
\(x-8:4-\left(46-23.2+6.3\right)=0\)
\(x-2-\left(46-46+18\right)=0\)
\(x-2-18=0\)
\(x-2=0+18\)
\(x-2=18\)
\(x=18+2\)
\(x=20\)
a) \(128-3\cdot\left(x+4\right)=23\)
\(\Leftrightarrow3\cdot\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
\(\Leftrightarrow x=31\)
b) \(\left(158-x\right):7=20\)
\(\Leftrightarrow158-x=140\)
\(\Leftrightarrow x=18\)
\(a,128-3.\left(x+4\right)=23\)
\(\Rightarrow3.\left(x+4\right)=128-23\)
\(\Rightarrow3.\left(x+4\right)=105\)
\(\Rightarrow\left(x+4\right)=105:3\)
\(\Rightarrow\left(x+4\right)=35\)
\(\Rightarrow x=35-4\)
\(\Rightarrow x=31\)
b) \(23-\frac{x}{25}=\frac{4}{5}\)
\(\frac{x}{25}=23-\frac{4}{5}\)
\(\frac{x}{25}=\frac{111}{5}=\frac{555}{25}\)
=> x = 555
\(5x+2x\cdot\left(2^3\cdot5-3^2\cdot4\right)+5^2=4^3\)
\(\Rightarrow5x+2x\cdot\left(8\cdot5-9\cdot4\right)+25=64\)
\(\Rightarrow5x+2x\cdot\left(40-36\right)=64-25\)
\(\Rightarrow5x+2x\cdot4=39\)
\(\Rightarrow5x+8x=39\)
\(\Rightarrow x\cdot\left(5+8\right)=39\)
\(\Rightarrow13x=39\)
\(\Rightarrow x=\dfrac{39}{13}\)
\(\Rightarrow x=3\)
Vậy: ...
\(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
x2.(x2 + 4) - x2 - 4=0
⇒ x2.(x2 + 4) - (x2 + 4) =0
⇒ (x2 + 4) .(x2 - 1) = 0
\(\Rightarrow\left[{}\begin{matrix}x^2+4=0\\x^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2=-4\\x^2=1\end{matrix}\right.\)(loại do x2 ≥ 0) \(\Rightarrow x=\pm1\)
a) x - 23 = 15 + 58
x - 23 = 73
x = 73 + 23
x = 96
b) x : 23 = 89 - 75
x : 23 = 14
x = 14 x 23
x = 322
a) x - 23 = 15 + 58
x - 23 = 73
x = 73 + 23
x = 96
b) x : 23 = 89 - 75
x : 23 = 14
x = 14 × 23
x = 322.
1. Ta có:
\(x^3-9x^2+27x-26=x^3-2x^2-7x^2+14x+13x-26\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)+13\left(x-2\right)=\left(x-2\right)\left(x^2-7x+13\right)\)
Thay x = 23, ta có: \(C=\left(23-2\right)\left(23^2-7.23+13\right)=8001\)
2.
a) \(x^2+4y^2+6x-12y+18=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-12y+9\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(2y-3\right)^2=0\)
Mà \(\left(x-3\right)^2\ge0\) với mọi x, \(\left(2y-3\right)^2\ge0\) với mọi y
\(\Rightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)và \(\left(2y-3\right)^2=0\Leftrightarrow2y-3=0\Leftrightarrow y=\frac{3}{2}\)
Vậy \(\left(x,y\right)=\left(3;\frac{3}{2}\right)\)
b) \(2x^2+2y^2+2xy-10x-8y+41=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2=0\)
.....................................
Rồi giải tương tự như trên
2 TH
TH1 : x + 4 = 0 => x = -4
TH2 : 23 - x = 0 => x = 23
vậy x = -4 và x = 23