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a) \(\Leftrightarrow\left(-63x^2+78x-15\right)+\left(63x^3+x-20\right)=44\)
\(\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\)
\(\Leftrightarrow79x-35=44\)
\(\Leftrightarrow79x=44+35\)
\(\Leftrightarrow79x=79\)
\(\Leftrightarrow x=1\)
b) \(\Leftrightarrow\left(x^2+3x+2\right).\left(x+5\right)-x^2.\left(x+8\right)=27\)
\(\Leftrightarrow x.\left(x^2+3x+2\right)+5.\left(x^2+3x+2\right)-x^3-8x^2=27\)
\(\Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\)
\(\Leftrightarrow17x+10=27\)
\(\Leftrightarrow17x=17\)
\(\Leftrightarrow x=1\)
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một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
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mk giải từng nha == tại vì mk sợ nhiều qus bị troll
\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(27x^3+18x^2+12x-18x^2-12x-8-3x\left(9x^2-3x+1\right)+\left(9x^2-3x+1\right)=x-4\)
\(27x^3-8-3\left(9x^2-3x+1\right)+9x^2-3x+1=x-4\)
\(27x^3-7-3x\left(9x^2-3x+1\right)+9x^2-3x=x-4\)
\(27x^3-7-27x^3+9x^2-3x+9x^2-3x=x-4\)
\(-7+18x^2-6x=x-4\)
\(3-18x^2+7x=0\)
\(x=\frac{-7+\sqrt{265}}{-36};\frac{-7-\sqrt{265}}{-36}\)
\(9\left(2x+1\right)=4\left(x-5\right)^2\)
\(18x+9=4x^2-40x+100\)
\(18x+9-4x^2+40x-100=0\)
\(58x-91-4x^2=0\)
\(x=\frac{29-3\sqrt{53}}{4};\frac{29+3\sqrt{53}}{4}\)
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Câu hỏi của Trịnh Minh Châu - Toán lớp 8 - Học toán với OnlineMath
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+) \(E=\left(4-x\right)\left(x+4\right)-\left(x+2\right)^2=16-x^2-x^2-4x-4\)
\(=-2x^2-4x+12\)
+)\(P=\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=4x^2+12x+9-4x^2+1\)
\(=12x+10\)
+)\(Q=\left(4-3x\right)^2-\left(9x-1\right)\left(9x+1\right)=16-24x+9x^2-81x^2+1\)
\(=-72x^2-24x+17\)
+) \(M=\left(5+x\right)\left(x-5\right)-\left(x-3\right)^2=x^2-25-x^2+6x-9\)
\(=6x-34\)
+) \(N=2\left(3x+1\right)\left(x-2\right)-6\left(x+2\right)^2=\left(6x+2\right)\left(x-2\right)-6\left(x^2+4x+4\right)\)
\(=6x^2-12x+2x-4-6x^2-24x-24=-34x-28\)
\(E=\left(4-x\right)\left(x+4\right)-\left(x+2\right)^2\)
\(E=4x+16-x^2-4x-x^2-4x-4\)
\(E=-2x^2-4x+12\)
\(P=\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)\)
\(P=4x^2+12x+9-4x^2-1\)
\(P=12x+10\)
\(Q=\left(4-3x\right)^2-\left(9x-1\right)\left(9x+1\right)\)
\(Q=16-24x+9x^2-81x^2+1\)
\(Q=17-24x-74x^2\)
\(M=\left(5+x\right)\left(x-5\right)-\left(x-3\right)^2\)
\(M=5x-25+x^2-5x-x^2+6x-9\)
\(M=-34+6x\)
\(N=2\left(3x+1\right)\left(x-2\right)-6\left(x+2\right)^2\)
\(N=\left(6x+2\right)\left(x-2\right)-\left(6x+12\right)^2\)
\(N=6x^2-12x+2x-4-36x^2-144x-144\)
\(N=-30x^2-154x-148\)
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)=\left(3x+2\right)\left(3x-2\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x+2\right)\left(3x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow3x+2=0\) ; \(x+1=0\) ; \(-2x+1=0\)
+) \(3x+2=0\)
\(\Leftrightarrow x=\dfrac{-2}{3}\)
+) \(x+1=0\)
\(\Leftrightarrow x=-1\)
+) \(-2x+1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Tập nghiệm: \(S=\left\{\dfrac{-2}{3};\dfrac{1}{2};-1\right\}\)
(3x+2)(x2-1)=(9x2-4)(x+1)
<=> (3x+2)(x+1)(x-1)-(3x-2)(3x+2)(x+1)=0
<=> (3x+2)(x+1)(x-1-3x+2)=0
<=> (3x+2)(x+1)(1-2x)=0
<=> \(\left\{{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy....\(\left\{{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=-2\\x=-1\\-2x=-1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)