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\(\dfrac{140}{x}+5=\dfrac{\left(140+10\right)}{x-1}\left(x\ne0,x\ne1\right)\)
\(\Leftrightarrow\dfrac{140+5x}{x}=\dfrac{150}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(140+5x\right)=150x\)
\(\Leftrightarrow140x+5x^2-140-5x-150x=0\)
\(\Leftrightarrow5x^2-15x-140=0\)
\(\Leftrightarrow x^2-3x-28=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(N\right)\\x=-4\left(N\right)\end{matrix}\right.\)
\(S=\left\{7,-4\right\}\)
ĐK: `x \ne 0 ; x \ne -1`
`140/x+5=150/(x-1)`
`<=>(140+5x)/x=150/(x-1)`
`<=>(140x+5x)(x-1)=150x`
`<=>5x^2+135x-140=150x`
`<=>5x^2-15x-140=0`
`<=>` \(\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy...
`(a+5)xx[(140/a)-1]=150`
`<=>(a+5).((140-a)/a)=150`
`<=>(a+5)(140-a)=150a`
`<=>140a-a^2+700-5a=150`
`<=>700+135a-a^2=150`
`<=>a^2-15a-700=0`
`Delta=225+2800=3025`
`<=>a_1=35,a_2=-20`
Vậy `S={35,-20}`.
=>\(\dfrac{140x+700-150x}{x\left(x+5\right)}=1\)
=>x^2+5x=-10x+700
=>x^2+15x-700=0
=>(x+35)(x-20)=0
=>x=20 hoặc x=-35
c) \(\sqrt{5+\sqrt{24}}=\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
d) \(\sqrt{12-\sqrt{140}}=\sqrt{12-2\sqrt{35}}=\sqrt{7}-\sqrt{5}\)
f) \(\sqrt{8-\sqrt{28}}=\sqrt{8-2\sqrt{7}}=\sqrt{7}-1\)
g) \(\sqrt{23-4\sqrt{15}}=\sqrt{23-2\cdot\sqrt{60}}=2\sqrt{5}-\sqrt{3}\)
h) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)
\(\left\{{}\begin{matrix}a.b=140\\\left(a+5\right)\left(b-1\right)=150\end{matrix}\right.< =>\left\{{}\begin{matrix}b=\dfrac{140}{a}\left(1\right)\\\left(a+5\right)\left(b-1\right)=150\left(2\right)\end{matrix}\right.\)(\(a\ne0\))
thế(1) vào(2)\(=>\left(a+5\right)\left(\dfrac{140}{a}-1\right)=150\)
\(< =>140-a+\dfrac{700}{a}-5=150\)
\(< =>\dfrac{700}{a}-a=15\)
\(< =>\dfrac{700-a^2}{a}=15=>-a^2+700=15a< =>-a^2-15a+700=0\)
\(\Delta=\left(-15\right)^2-4\left(-1\right)700=3025>0\)
\(=>\left[{}\begin{matrix}a1=\dfrac{15+\sqrt{3025}}{2\left(-1\right)}=-35\left(TM\right)\\a2=\dfrac{15-\sqrt{3025}}{2\left(-1\right)}=20\left(TM\right)\end{matrix}\right.\)
với a=a1=-35 thay vào(1)=>\(b=\dfrac{140}{-35}=-4\)
với a=a2=20 tahy vào (1)=>\(b=\dfrac{140}{20}=7\)
Vậy(a,b)={(-35;-4);(20;7)}