* ĐỀ như dưới phải ko ??? Ai lại ghi x2 bao giờ ??

a) (x + 8)(x + 6) - x² =104

\(\Leftrightarrow\) x² + 6x + 8x + 48 - x² = 104

\(\Leftrightarrow\) 14x + 48 = 104

\(\Leftrightarrow\) 14x = 56

\(\Leftrightarrow\) x = 4

a: \(=\left(x-y\right)\left(x+y\right)\)

\(=74\cdot100=7400\)

c: \(=\left(x+2\right)^3\)

\(=10^3=1000\)

a) \(=\left(x-y\right)\left(x+y\right)\)

    Thay \(x=87;y=13\) ta đc:   \(\left(87-13\right)\left(87+13\right)=74\cdot100=7400\)

b)\(=\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)

   Thay \(x=10;y=-1\) ta đc:

    \(10^3-\left(-1\right)^3=1000-1=999\)

c)\(=\left(x+2\right)^3\)

   Thay \(x=8\) ta đc: \(\left(8+2\right)^3=10^3=1000\)

d)\(=x^2-8x+16+1=\left(x-4\right)^2+1\)

   Thay \(x=104\) ta đc: \(\left(104-4\right)^2+1=100^2+1=10001\)

Ta có:

(x+8)(x+6)-x^2=104

x^2+6x+8x+48-x^2=104

14x+48=104

14x=104-48

14x=56

x=4

(x+8)(x+6)-x²=104
<=> x²+6x+8x+48-x²=104
<=> 14x+48=104
<=> 14x=104-48
<=> 14x=56
<=> x= 56:14
<=> x=4

\(\frac{x+110}{100}+\frac{x+8}{102}+\frac{x+6}{104}=\frac{x+4}{106}+\frac{x+2}{108}\)

\(\Leftrightarrow\frac{x+110}{100}+\left(\frac{x+8}{102}+1\right)+\left(\frac{x+6}{104}+1\right)=\left(\frac{x+4}{106}+1\right)+\left(\frac{x+2}{108}+1\right)\)

\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}=\frac{x+110}{106}+\frac{x+110}{108}\)

\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}-\frac{x+110}{106}-\frac{x+110}{108}=0\)

\(\Leftrightarrow\left(x+110\right)\left(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}+\frac{1}{106}+\frac{1}{108}\right)=0\)

\(\Leftrightarrow x+110=0\) (vì \(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}-\frac{1}{106}-\frac{1}{108}>0\))

\(\Leftrightarrow x=-110\)

Vậy \(x=-110\)

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a) Ta có: 

\(X+6=8\)

\(\Rightarrow X=8-6\)

\(\Rightarrow X=2\)

Vậy: \(A=\left\{2\right\}\)

Có 1 phần tử 

b) Số phần tử của tập hợp B là:

\(\left(104-2\right):2+1=52\) (số hạng)

có 52 số hạng

\(x+\frac{5}{7}=\frac{7}{8}:\frac{1}{6}\)

\(\Leftrightarrow x+\frac{5}{7}=\frac{21}{4}\)

\(\Leftrightarrow x=\frac{127}{28}\)

\(\frac{72}{104}=\frac{72:8}{104:8}=\frac{9}{13}\)

\(\frac{3}{13}< \frac{4}{13}< \frac{5}{13}< \frac{6}{13}< \frac{7}{13}< \frac{8}{13}< \frac{72}{104}=\frac{9}{13}\)

Tìm x :

a ) \(x+\frac{5}{7}=\frac{7}{8}\div\frac{1}{6}=\frac{7\times6}{8\times1}=\frac{21}{4}\)

\(x=\frac{21}{4}-\frac{5}{7}=\frac{127}{28}\)

b ) \(\frac{15}{6}-x=\frac{5}{8}\times\frac{2}{15}=\frac{1}{12}\)

\(x=\frac{15}{6}-\frac{1}{12}=\frac{29}{12}\)

Rút gọn : \(\frac{72}{104}=\frac{72\div8}{104\div8}=\frac{9}{13}\)

Điền STN thích hợp :

\(\frac{3}{13}< \frac{4}{13}< \frac{5}{13}< \frac{6}{13}< \frac{7}{13}< \frac{8}{13}< \frac{9}{13}\left(\frac{72}{104}\right)\)Những số cần điền là : 4, 5, 6, 7, 8 .

==========> CHÚC HỌC GIỎI  NHỚ K MÌNH <==========

\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)

\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)

c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)

\(\Rightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)