A,3.x - 16 : 2^3= 31; B,2^10 : 2^8 + 3. [4. 7 + 3.4 ] ; C,4^6 : 4^3 - 2^2 . 2^3; D,141 + 2^5 . 2^4 - 3^1 . 3^2 ; E,x + 2^5 : 2^4 = 4.4^2 giúp mình với
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\(a=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)
\(\Leftrightarrow a^3=16-8\sqrt{5}+16+8\sqrt{5}+3\sqrt[3]{\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)}\cdot a\)
\(\Leftrightarrow a^3=32+3\sqrt[3]{256-320}\cdot a\)
\(\Leftrightarrow a^3=32-12a\)
Giải pt được \(a=2\).
Khi đó : \(P\left(a\right)=\left(2^2+12\cdot2-31\right)=-3\)
Vậy...
a) => 6x-6-5x=15
=> 6x-6+6-5x=15+6
=> 6x-5x=21
=> x=21
b) => 8x-4-7x=16
=> 8x-4+4-7x=16+4
=> 8x-7x=20
=> x=20
c) => 8-[-(2.4x)]-9x-21=31
=> 8-(-8x)-9x=31+21
=> 8-(-8x)-9x=52
=> (-8x)-9x=8-52
=> (-8x)+(-9x)=-44
=> -17x=-44
=> x=(-44):(-17)
=> x=
câu c chắc tớ làm sai
a) \(\dfrac{-15+9+11}{16}=\dfrac{5}{16}\)
b) \(\dfrac{2}{3}\left(1,4+1,6-1,2\right)=\dfrac{2}{3}\times\dfrac{9}{5}=\dfrac{6}{5}\)
c) \(3\dfrac{2}{15}\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{31}{15}=\dfrac{47}{15}-\dfrac{31}{15}=\dfrac{16}{15}\)
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
a: \(\dfrac{96}{\left(x-4\right)\left(x+4\right)}+\dfrac{7+x}{4-x}=\dfrac{2x-1}{x+4}-3\)
\(\Leftrightarrow\dfrac{96}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(x+7\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{3\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(96-x^2-11x-28=2x^2-9x+4-3\left(x^2-16\right)\)
\(\Leftrightarrow-x^2-11x+68=2x^2-9x+4-3x^2+48\)
\(\Leftrightarrow-x^2-11x+68=-x^2-9x+52\)
=>-11x+68=-9x+52
=>-2x=-16
hay x=8(nhận)
b: \(\dfrac{2}{x-1}+\dfrac{3}{x-2}=\dfrac{3}{x-3}\)
\(\Leftrightarrow2\left(x-2\right)\left(x-3\right)+3\left(x-1\right)\left(x-3\right)=3\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(x^2-5x+6\right)+3\left(x^2-4x+3\right)=3\left(x^2-3x+2\right)\)
\(\Leftrightarrow2x^2-10x+12+3x^2-12x+9=3x^2-9x+6\)
\(\Leftrightarrow5x^2-22x+21-3x^2+9x-6=0\)
\(\Leftrightarrow2x^2-13x+15=0\)
\(\Leftrightarrow2x^2-10x-3x+15=0\)
=>(x-5)(2x-3)=0
=>x=5(nhận) hoặc x=3/2(nhận)
`3x-16:2^3=31`
`=>3x-16:8=31`
`=>3x-2=31`
`=>3x=31+2`
`=>3x=33`
`=>x=11`
__
`2^10:2^8+3[4.7+3.4]`
`=2^2+3[4(3+7)]`
`=4+3[4.10]`
`=4+3.40`
`=4+120`
`=124`
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`4^6:4^3-2^2 . 2^3`
`=4^3-2^5`
`=64-32`
`=32`
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`141+2^5 . 2^4-3^1 . 3^2`
`=141+2^9-3^3`
`=141+512-9`
`=644`
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`x+2^5:2^4=4.4^2`
`=>x+2=4^3`
`=>x=64-2`
`=>x=62`