240-[23+(13+24.3-x)]=132

240-[23+(13+168-x)]=132

240-[23+(181-x)]=132

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x-8:4-(46-23.2+6.3)=0

\(x-8:4-\left(46-23.2+6.3\right)=0\)

\(x-2-\left(46-46+18\right)=0\)

\(x-2-18=0\)

\(x-2=0+18\)

\(x-2=18\)

\(x=18+2\)

\(x=20\)

\(240-\left[23+\left(13+24.3-x\right)\right]=132\)

\(240\left[23+\left(13+82-x\right)\right]=132\)

\(23+\left(95-x\right)=132:240\)

\(23+\left(95-x\right)=\frac{132}{240}=\frac{11}{30}\)

\(23+\left(95-x\right)=\frac{11}{30}\)

\(95-x=\frac{11}{30}-23\)

\(95-x=\frac{11}{30}-\frac{690}{30}\)

\(95-x=-\frac{679}{30}\)

\(x=95+\frac{679}{30}\)

\(x=\frac{2850}{30}+\frac{679}{30}\)

\(x=\frac{3529}{30}=\)tự rút gọn ( nếu có thể )

a) \(128-3\cdot\left(x+4\right)=23\)

\(\Leftrightarrow3\cdot\left(x+4\right)=105\)

\(\Leftrightarrow x+4=35\)

\(\Leftrightarrow x=31\)

b) \(\left(158-x\right):7=20\)

\(\Leftrightarrow158-x=140\)

\(\Leftrightarrow x=18\)

\(a,128-3.\left(x+4\right)=23\)                                                      

\(\Rightarrow3.\left(x+4\right)=128-23\)

\(\Rightarrow3.\left(x+4\right)=105\)

\(\Rightarrow\left(x+4\right)=105:3\)

\(\Rightarrow\left(x+4\right)=35\)

\(\Rightarrow x=35-4\)

\(\Rightarrow x=31\)

\(b,\left(158-x\right):7=20\)

\(\Rightarrow\left(158-x\right)=20.7\)

\(\Rightarrow\left(158-x\right)=140\)

\(\Rightarrow x=158-140\)

\(\Rightarrow x=18\)

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

Mình làm cho bạn sao bạn không tick vậy 

GTLN A = -19

Đố bn bit tai sao? nêu k bit mk giai dùm

\(5x+2x\cdot\left(2^3\cdot5-3^2\cdot4\right)+5^2=4^3\)

\(\Rightarrow5x+2x\cdot\left(8\cdot5-9\cdot4\right)+25=64\)

\(\Rightarrow5x+2x\cdot\left(40-36\right)=64-25\)

\(\Rightarrow5x+2x\cdot4=39\)

\(\Rightarrow5x+8x=39\)

\(\Rightarrow x\cdot\left(5+8\right)=39\)

\(\Rightarrow13x=39\)

\(\Rightarrow x=\dfrac{39}{13}\)

\(\Rightarrow x=3\)

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.

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

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) \(2x^2+2y^2+2xy-10x-8y+41=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2=0\)

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Rồi giải tương tự như trên