40x4-26x3+31x2
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\(26\times3+15\times3+4\times2-3\times2\)
\(=3\times\left(26+15\right)+2\times\left(4-3\right)\)
\(=3\times41+2\times1\)
\(=123+2\)
\(=125\)
26\(\times\)3 + 15\(\times\)3 + 4\(\times\)2 - 3\(\times\)2
= (26\(\times\)3 + 14\(\times\)3) + 3 + (4\(\times\)2 - 3\(\times\)2)
= 3\(\times\)(26 +14) + 3 + 2\(\times\)(4 - 3)
= 3 \(\times\)40 + 3 + 2
= 120 + 5
= 125
1)
`7x^2 -49x=0`
`<=>x(7x-49)=0`
\(< =>\left[{}\begin{matrix}x=0\\7x-49=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
2)
`8x^2 -16x=0`
`<=>x(8x-16)=0`
\(< =>\left[{}\begin{matrix}x=0\\8x-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
3)
`2x^3 +40x=0`
`<=>x(2x^2 +40)=0`
`<=>x=0` hoặc`2x^2 +40=0`
`<=>x=0` hoặc `2x^2 =-40` (vô lí vì `2x^2` luôn lớn hơn hoặc bằng 0)
`<=>x=0`
4)
`-x^3 +16x=0`
`<=>x^3 -16x=0`
`<=>x(x^2 -16)=0`
\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
a: Ta có: \(40x^4+5x=0\)
\(\Leftrightarrow5x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(8x^2-2x-1=0\)
\(\Leftrightarrow8x^2-4x+2x-1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
26X2+26X3+26=
26X2+26X3+26X1=
26X(1+2+3)=
26X6=
156
nhớ k cho mình nhé
thank you
\(2x+x:3-20=13.\left(98-26.3\right)\)
\(\Rightarrow2x+\dfrac{1}{3}x-20=13.\left(98-78\right)\)
\(\Rightarrow\dfrac{7}{3}x=13.20+20\)
\(\Rightarrow\dfrac{7}{3}x=20.\left(13+1\right)\)
\(\Rightarrow\dfrac{7}{3}x=20.14\Rightarrow\dfrac{7}{3}x=280\)
\(\Rightarrow x=280:\dfrac{7}{3}=120\)
Chúc bạn học tốt!!!
\(2x+x:3-20=13(98-26.3)\)
\(2x+x:3-20=13.20\)
\(2x+x:3-20=260\)
\(2x+x:3=280\)
\(x\left(2+\dfrac{1}{3}\right)=280\)
\(\dfrac{7}{3}x=280\)
\(x=120\)
a) \(40x^4-10x^2=10x^2\left(4x^2-1\right)=10x^2\left(2x-1\right)\left(2x+1\right)\)
b) \(16x^4-20x^2-y^2-5y=\left(4x^2-\dfrac{5}{2}\right)^2-\left(y-\dfrac{5}{2}\right)^2=\left(4x^2-\dfrac{5}{2}-y+\dfrac{5}{2}\right)\left(4x^2-\dfrac{5}{2}+y-\dfrac{5}{2}\right)=\left(4x^2-y\right)\left(4x^2+y-5\right)\)c)\(64a^2-9b^2-16a+1=\left(8a-1\right)^2-9b^2=\left(8a-1-3b\right)\left(8a-1+3b\right)\)d) \(5x^2+23x-10=5\left(x-\dfrac{2}{5}\right)\left(x+5\right)\)
a: \(40x^4-10x^2\)
\(=10x^2\left(4x^2-1\right)\)
\(=10x^2\cdot\left(2x-1\right)\left(2x+1\right)\)
b: \(16x^4-20x^2-y^2-5y\)
\(=\left(4x^2-y\right)\left(4x^2+y\right)-5\left(4x^2+y\right)\)
\(=\left(4x^2+y\right)\left(4x^2-y-5\right)\)
c: Ta có: \(64a^2-9b^2-16a+1\)
\(=\left(8a-1\right)^2-9b^2\)
\(=\left(8a-1-3b\right)\left(8a-1+3b\right)\)
d: Ta có: \(5x^2+23x-10\)
\(=5x^2+25x-2x-10\)
\(=\left(x+5\right)\left(5x-2\right)\)
40 x 4 - 26 x 3 + 31 x 3 = 160 - 78 + 62 =144
= 160 + 78 + 62
= 300 nha bn