12x+4/21-x-3/3=3(x-2)/7
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*)\(b^2+c^2=a^2\)
\(\Leftrightarrow b^2=a^2-c^2\)
\(\Leftrightarrow b=\sqrt{a^2-c^2}\)
Ta có: \(\sqrt{a^2-c^2}>c\Leftrightarrow a^2-c^2>c^2\)
\(\Leftrightarrow a^2>2c^2\)(luôn đúng)
=> c<b
*) \(a^2=b^2+c^2\Leftrightarrow\hept{\begin{cases}c=3\\b=4\\a=5\end{cases}\Leftrightarrow c=b+1}\)
1) x2-1=0 <=> x2=1 <=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
2) x2+1=0
<=> x2=-1
Mà x2 >=0 với mọi x; -1<0
=> không có x thỏa mãn
3) \(x^3-x^2-21x+45=0\)
\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)
<=> x2(x-3) +2x(x-3)-15(x-3)=0
<=> (x-3)(x2 +2x-15)=0
<=> (x-3)(x2+5x-3x-15)=0
<=> (x-3)[x(x+5)-3(x+5)]=0
<=> (x-3)2(x+5)=0
\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)^2\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-5\end{cases}}}\)
\(3x\left(x+5\right)-\left(x+2\right)^2=2x^2+7\)
\(\Leftrightarrow3x^2+15x-x^2-4x-4=2x^2+7\)
\(\Leftrightarrow3x^2-2x^2-x^2+15x-4x=7+4\)
\(\Leftrightarrow11x=11\)
\(\Leftrightarrow x=1\)
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đăng kí hộ
\(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)
<=> \(\frac{3x+2}{x-1}+\frac{2\left(x-2\right)}{x+2}=5\)
<=> (3x + 2)(x + 2) + 2(x - 2)(x - 1) = 5(x - 1)(x + 2)
<=> 3x2 + 6x + 2x + 4 + 2x2 - 2x - 4x + 4 = 5x2 + 10x - 5x - 10
<=> 5x2 + 2x + 8 = 5x2 + 5x - 10
<=> 5x2 + 2x + 8 - 5x2 = 5x - 10
<=> 2x + 8 = 5x - 10
<=> 2x + 8 - 5x = -10
<=> -3x + 8 = -10
<=> -3x = -10 - 8
<=> -3x = -18
<=> x = 6
\(3\left(x-1\right)+4\left(2x-1\right)=-8\)
\(\Leftrightarrow3x-3+8x-4=-8\)
\(\Leftrightarrow3x+8x=-8+4+3\)
\(\Leftrightarrow11x=-1\)
\(\Leftrightarrow x=-\frac{1}{11}\)
3(x - 1) + 4(2x - 1) = -8
<=> 3x - 3 + 8x - 4 = -8
<=> 11x = -8 + 4 +3
<=> 11x = -1
<=> x = \(\frac{-1}{11}\)
(12x + 4)/21 - (x - 3)/3 = (3(x - 2))/7
<=> (4(3x + 1))/21 - (x - 3)/3 = (3(x - 2))/7
<=> 4(3x + 1) - 7(x - 3) = 9(x - 2)
<=> 12x + 4 - 7x + 21 = 9x - 18
<=> 5x + 25 = 9x - 18
<=> 5x + 25 - 9x = -18
<=> -4x + 25 = -18
<=> -4x = -18 - 25
<=> -4x = -43
<=> x = 43/4