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\(\frac{2x+1}{89}+\frac{2x+4}{86}+\frac{2x+8}{82}+3=0\)
\(\frac{2x+1}{89}+1+\frac{2x+4}{86}+1+\frac{2x+8}{82}+1-3+3=0\)
\(\frac{2x+90}{89}+\frac{2x+90}{86}+\frac{2x+90}{82}=0\)
\(\left(2x+90\right)\left(\frac{1}{89}+\frac{1}{86}+\frac{1}{82}\right)=0\)
mà \(\frac{1}{89}+\frac{1}{86}+\frac{1}{82}\ne0\)
\(\Rightarrow2x+90=0\)
\(\Rightarrow2x=-90\)
\(\Rightarrow x=-45\)
Vậy \(x=-45\)
@Lam Ngo Tung dòng 2 công mỗi phân thức thêm 1 rồi trừ đi 3 sao cộng tiếp với 3 thế :v
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a: \(\Leftrightarrow2x^3-56-7x^2+16x=0\)
\(\Leftrightarrow2x\left(x^2+8\right)-7\left(x^2+8\right)=0\)
=>2x-7=0
hay x=7/2
b: \(\Leftrightarrow x^5\left(x^2+2\right)+x\left(x^2+2\right)=0\)
=>x(x2+2)(x4+1)=0
=>x=0
c: \(\Leftrightarrow2x^2+x-5x-\dfrac{5}{2}=0\)
\(\Leftrightarrow2x^2-4x-\dfrac{5}{2}=0\)
hay \(x\in\left\{\dfrac{5}{2};-\dfrac{1}{2}\right\}\)
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x - 212 - 2x + 13 = 3x + 56 - x
x - 2x - 3x + x = 212 - 13 + 56
-3x = 255
x = - 85
theo đề bài, ta có:
x-212-2x+13=3x+56-x
\(\Rightarrow\)x-2x-3x+x=212+56-13
\(\Rightarrow\)-3x=255
\(\Rightarrow\)x=\(\frac{255}{-3}\)
\(\Rightarrow\)x=-85
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a) 27x^3 –27x^2 +18x –4
= 27x^3 –9x^2–18x^2+6x + 12x –4
= 9x^2 (3x–1) – 6x (3x–1) +4(3x–1)
= (3x-1) (9x^2–6x+4)
b)2x^3–2x^2+5x+3
= 2x^3+x^2–2x^2–x+6+3
= x^2(2x+1)-x^2(2x+1)+3(2x+1)
= (2x+1) 3
c) 2x^4 + 5x^3+13x^2+25x+15
=2x^3(x+1)+3x^2(x+1)+10x(x+1)+15(x+1)
=(x+1)(x^2(2x+3)+5(2x+3))
=(x+1)(2x+3)(x^2+5)
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E đây cx ngu Toán,sai thì thoy đừng ném đá
(x+3)^3 - (x+1)^3=56
<=>x^3 + 9.x^2+27x+27-x^3-3.x^2-3x-1=56
<=>6.x^2+24x+26=56
<=>6.x^2+24x-3=0
<=>6x^2-6x+30x=0
<=>6x(x-1)+30(x-1)=0
<=>(x-1)(6x+30)=0
<=>x=\(\hept{\begin{cases}1\\-5\end{cases}}\)
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Ta có:
\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-\left(ab+bc+ca\right)+\left(a+b+c\right)-1\)
\(=abc-\left(ab+bc+ca\right)+5\ge0\)
\(\Rightarrow abc\ge ab+bc+ca-5\)(1)
\(\left(a-3\right)\left(b-3\right)\left(c-3\right)=abc-3\left(ab+bc+ca\right)+9\left(a+b+c\right)-27\)
\(=abc-3\left(ab+bc+ca\right)+27\le0\)
\(\Rightarrow abc\le3\left(ab+bc+ca\right)-27\)(2)
(1)(2) suy ra \(ab+bc+ca-5\le3\left(ab+bc+ca\right)-27\)
\(\Leftrightarrow ab+bc+ca\ge11\).
\(6^2=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2=36-2\left(ab+bc+ca\right)\le36-2.11=14\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\\\left(a-3\right)\left(b-3\right)\left(c-3\right)=0\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}}\)và các hoán vị.
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\(336x^2\) - \(56xy\) = \(50,4x+100,8x\)
\(336x^2-56xy-151,2x=0\)
\(56\left(6x^2-xy-2,7x\right)=0\)
\(6x-y-2,7=0\)
\(x=\frac{2,7+y}{6}\)
2x=86-56
2x=30
x=15