(1 -x )^3+158 =131
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a) 1. = (-75-25) + (231-131)
= -100 + 100 = 0
2. 45/9 + 9x(-1) +1
= 5- 9 +1 = -3
3. 6. (-83) + 6.(-37) + 20.6
= 6 ( 20-83-37)
= 6. (-100) = -600
4. = 26.76 - 26.38 - 76.26 +76 .38
= 38 ( 76-26)
= 38.50 = 1900
3. 3n+2 = 3n-3 + 5 = 3(n-1) + 5
3(n-1) chia hết cho n-1 nên để 3n+2 chia hết cho n-1 thì 5 chia hết cho n-1
=> n-1 thuộc tập cộng trừ 1; 5
kẻ bảng => n = 2; 0; 6; -4
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`2x -2/3 +1/2x =-1`
`=> 2x+1/2x =-1+2/3`
`=> (2+1/2) x =-3/3 + 2/3`
`= 5/2 x = -1/3`
`=> x=-1/3 : 5/2`
`=>x= -2/15`
`31/36 - (1/3-x)^2 =5/6`
`=> (1/3-x)^2 = 31/36 - 5/6`
`=> (1/3-x)^2 =1/36`
`=> (1/3-x)^2 = (+-1/6)^2`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}-x=\dfrac{1}{6}\\\dfrac{1}{3}-x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{1}{2}\end{matrix}\right.\)
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\(\sqrt[3]{x+1}=x^3-15x^2+75x-125-6=0\)
\(\Leftrightarrow\sqrt[3]{x+1}+6=\left(x-5\right)^3\)
Đặt \(\sqrt[3]{x+1}=a-5\) ta được hệ:
\(\left\{{}\begin{matrix}\left(a-5\right)^3=x+1\\a-5+6=\left(x-5\right)^3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a-5\right)^3=x+1\\\left(x-5\right)^3=a+1\end{matrix}\right.\)
Trừ vế cho vế ta được:
\(\left(x-5\right)^3-\left(a-5\right)^3=a-x\)
\(\Leftrightarrow\left(x-a\right)\left(\left(x-5\right)^2+\left(x-5\right)\left(a-5\right)+\left(a-5\right)^2\right)+\left(x-a\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left[\left(x-5+\frac{a-5}{2}\right)^2+\frac{3\left(a-5\right)^2}{4}+1\right]=0\)
\(\Leftrightarrow x-a=0\) (phần ngoạc phía sau luôn dương)
\(\Leftrightarrow x=a\Leftrightarrow x=\sqrt[3]{x+1}+5\Leftrightarrow x-5=\sqrt[3]{x+1}\)
\(\Leftrightarrow x^3-15x^2+75x-125=x+1\)
\(\Leftrightarrow x^3-15x^2+74x-126=0\)
\(\Rightarrow x=7\)
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à mk nhầm, cái này mới đúng:
=1,75.(−1621)−(133+2,25):15860G=1,75.(−1621)−(133+2,25):15860
=−4,3−7912:7930G=−4,3−7912:7930
=−43−52G=−43−52
=−236
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Ta có \(131x-941=16\)
\(\Rightarrow131x=957\)
\(\Rightarrow x=\frac{957}{131}\)
b.\(12.\left(x-1\right):3=103\)
\(\Rightarrow12.\left(x-1\right)=309\)
\(\Rightarrow x-1=\frac{103}{4}\)
\(\Rightarrow x=\frac{107}{4}\)
(1 - x)3 + 158 = 131
=> (1 - x)3 = 131 - 158
=> (1 - x)3 = -27
=> (1 - x)3 = (-3)3
=> 1 - x = -3
=> x = 1 + 3
=> x = 4
Vậy ...
(1 - x)3 + 158 = 131
(1 - x)3 = 131 - 158
(1 - x)3 = -27
(1 - x)3 = -33
1 - x = -3
- x = -3 - 1
- x = -4
=> x = 4