15 x 2 + 15 x3 - 15 + 15 x 5
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\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
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\(TH_1:x\ge0\Leftrightarrow x^3\ge0\Leftrightarrow VT>0\left(loại\right)\)
\(TH_2:x< 0\)
Với \(x=-1\Leftrightarrow VT=4\cdot9\cdot14\cdot29>0\left(loại\right)\)
Với \(x=-2\Leftrightarrow VT=-3\cdot2\cdot7\cdot23< 0\left(nhận\right)\)
Với \(x=-3\Leftrightarrow VT=-22\left(-17\right)\left(-12\right)\cdot3< 0\left(nhận\right)\)
Với \(x< -4\Leftrightarrow x^3< -64\Leftrightarrow x^3+5< x^3+10< x^3+15< x^3+30< 0\)
Do đó cả 4 thừa số trong tích đều âm nên tích này luôn dương
Vậy \(x\in\left\{-2;-3\right\}\)
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a) Cách 1
\(\frac{8}{15}:\frac{2}{3}+\frac{7}{15}:\frac{2}{3}\)
\(=\frac{8}{15}.\frac{3}{2}+\frac{7}{15}.\frac{3}{2}\)
\(=\frac{4}{5}+\frac{7}{10}=\frac{3}{2}\)
Cách 2
\(\frac{8}{15}:\frac{2}{3}+\frac{7}{15}:\frac{2}{3}\)
\(=\left(\frac{8}{15}+\frac{7}{15}\right).\frac{3}{2}\)
\(=1.\frac{3}{2}=\frac{3}{2}\)
a) C1: 8/15 : 2/3 + 7/15 : 2/3
= 4/5 + 7/10
= 3/2
C2: 8/15 : 2/3 + 7/15 : 2/3
= ( 8/15 + 7/15 ) : 2/3
= 1 : 2/3
= 3/2
b) C1: 4/9 x 3/5 + 5/9 x 3/5
= 4/15 + 1/3
= 3/5
C2: 4/9 x 3/5 + 5/9 x 3/5
= ( 4/9 + 5/9 ) x 3/5
= 1 x 3/5
= 3/5
Chúc bạn học tốt :)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta thấy \(xy=\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}=\dfrac{3^2-5}{2}=2\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\) \(=3\left(5-2\right)=9\)
b) Ta thấy \(xy=\dfrac{-\left(x-y\right)^2+\left(x^2+y^2\right)}{2}=\dfrac{15-5^2}{2}=-5\)
\(\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)\) \(=5\left(15-5\right)=50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(x^3+\left(x-5\right)\left(x+8\right)=2x^2-37\\ \Leftrightarrow x^3+x^2+3x-40=2x^2-37\\ \Leftrightarrow x^3-x^2+3x-3=0\\ \Leftrightarrow x^2\left(x-3\right)+3\left(x-3\right)=0\\ \Leftrightarrow\left(x^2+3\right)\left(x-3\right)=0\)
Vì \(x^2+3\ge3>0\Rightarrow x-3=0\\ \Leftrightarrow x=3\)
b)
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\\ \Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\\ \Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt \(x^2+x=y\)
\(\Rightarrow y\left(y-2\right)=24\\ \Leftrightarrow y^2-2y+1=25\\ \Leftrightarrow\left(y-1\right)^2=25\\ \Leftrightarrow\left[{}\begin{matrix}y-1=5\\y-1=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}y=6\\y=-4\end{matrix}\right.\)
Nếu y = 6
\(\Rightarrow x^2+x=6\\ \Leftrightarrow x^2+x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Nếu y = -4
\(\Rightarrow x^2+x=-4\\ \Leftrightarrow x^2+x+\dfrac{1}{4}=-4+\dfrac{1}{4}\\ \Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=-\dfrac{15}{4}\)
Mà \(\left(x+\dfrac{1}{.2}\right)^2\ge0>-\dfrac{15}{4}\)
`=> Loại`
c) Vế còn lại là bao nhiêu?
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right).3=15\\ =>x-1=5\\ =>x=6\)
\(20:\left(x+1\right)-3=2\\ =>20:\left(x+1\right)=5\\ =>x+1=4\\ =>x=3\)
\(\left(x-1\right).3=15\)
\(x-1=15:3\)
\(x-1=5\)
\(x=5+1\)
\(x=6\)
Vậy \(x=6\)
\(20:\left(x+1\right)-3=2\)
\(\left(x+1\right)-3=20:2\)
\(\left(x+1\right)-3=10\)
\(x+1=10-3\)
\(x+1=7\)
\(x=7-1\)
\(x=6\)
Vậy \(x=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
194x6x5-10x60x3-68x15
=30x194-30x60-2x34x15
=30x194-30x60-30x34
30x(194-60-34)
=30x100
=3000
15 x 2 + 15 x3 - 15 + 15 x 5=30+45-15+75=135