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c) ĐKXĐ : \(x\ne4\)
Để biểu thức \(\frac{3x^3-4x^2+x-1}{x-4}\) nguyên với \(x\) nguyên thì :
\(3x^3-4x^2+x-1⋮x-4\)
\(\Leftrightarrow3x^3-12x^2+8x^2-32x+33x-132+131⋮x-4\)
\(\Leftrightarrow3x^2.\left(x-4\right)+8x.\left(x-4\right)+31.\left(x-4\right)+131⋮x-4\)
\(\Leftrightarrow131⋮x-4\)
\(\Leftrightarrow x-4\inƯ\left(131\right)\)
\(\Leftrightarrow x-4\in\left\{-1,1,131,-131\right\}\)
\(\Leftrightarrow x\in\left\{3,5,135,-127\right\}\)
d) ĐKXĐ : \(x\ne-\frac{3}{2}\)
Để biểu thức \(\frac{3x^2-x+1}{3x+2}\) nhận giá trị nguyên với \(x\) nguyên thì :
\(3x^2-x+1⋮3x+2\)
\(\Leftrightarrow3x^2+2x-3x-2+3⋮3x+2\)
\(\Leftrightarrow x.\left(3x+2\right)-\left(3x+2\right)+3⋮3x+2\)
\(\Leftrightarrow3⋮3x+2\)
\(\Leftrightarrow3x+2\inƯ\left(3\right)\)
\(\Leftrightarrow3x+2\in\left\{-1,1,-3,3\right\}\)
\(\Leftrightarrow x\in\left\{-1,-\frac{1}{3},-\frac{5}{3},\frac{1}{3}\right\}\) mà \(x\) nguyên
\(\Rightarrow x=-1\)
a) ĐKXĐ : \(x\ne3\)
Để \(\frac{2}{x-3}\)nguyên
=> \(2⋮x-3\)
=> \(x-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x-3 | 1 | -1 | 2 | -2 |
| x | 4 | 2 | 5 | 1 |
Cả 4 giá trị đều tmđk
KL : Vậy x = { 4 ; 2 ; 5 ; 1 }
b) ĐKXĐ : \(x\ne-2\)
Để \(\frac{3}{x+2}\)nguyên
=> \(3⋮x+2\)
=> \(x+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x+2 | 1 | -1 | 3 | -3 |
| x | -1 | -3 | 1 | -5 |
Cả 4 giá trị đều tmđk
KL : Vậy x = { -1 ; -3 ; 1 ; -5 }
a, ĐKXĐ : \(x\ne3\)
\(\frac{2}{x-3}\)có giá trị nguyên
\(\Leftrightarrow x-3\inƯ\left(2\right)\)
\(Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)
+, TH1 : \(x-3=1\Leftrightarrow x=4\left(TM\right)\)
+, TH2 : \(x-3=-1\Leftrightarrow x=2\left(TM\right)\)
+, TH3 : \(x-3=2\Leftrightarrow x=5\left(TM\right)\)
+, TH4 : \(x-3=-2\Leftrightarrow x=1\left(TM\right)\)
Vậy với \(x\in\left\{4;2;5;1\right\}\)thì \(\frac{2}{x-3}\)có giá trị nguyên
a, 15x3 - 15x = 0
15x(x2-1)=0
15x=0 hoặc x2-1=0 (tự tính nhoa)
b,3x2-6x+3=0
3(x2-2x+1)=0
x2 -2x+1=0:3=3
x2-2x=3-1=2
x(x-2)=0
x=0 hoặc x-2=0 (tự tính nhoa)
Bài làm
a) 15x3-15x=0
<=> 15x( x2 - 1 ) = 0
<=> \(\orbr{\begin{cases}15x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)
Vậy x = { 0; + 1 }
b) 3x2 - 6x + 3 = 0
<=> 3( x2 - 2x + 1 ) = 0
<=> x2 - 2x + 1 = 0
<=> ( x - 1 )2 = 0
<=> x - 1 = 0
<=> x = 1
Vậy x = 1
c) 5(x - 1) - 3x(1 - x) = 0
<=> 5(x - 1) + 3x(x - 1) = 0
<=> (5 + 3x)(x - 1) = 0
<=> \(\orbr{\begin{cases}5+3x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=1\end{cases}}}\)
Vậy x = { -5/3; 1 }
e) -7(x + 2) = 2x(x + 2)
<=> -7(x + 2 ) - 2x( x + 2 ) = 0
<=> (x + 2)(-7 - 2x) = 0
<=> \(\orbr{\begin{cases}x+2=0\\-7-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{7}{2}\end{cases}}}\)
Vậy x = { -2; x = -7/2 }
f)(2x - 3)(3x + 5) = (x - 1)(3x + 5)
<=> (2x - 3)(3x + 5) - (x - 1)(3x + 5) = 0
<=> (3x + 5)(2x - 3 - x + 1) = 0
<=> (3x + 5)(x - 2) = 0
<=> \(\orbr{\begin{cases}3x+5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=2\end{cases}}}\)
Vậy x = { -5/3; 2 }
a, \(2\left(x-1\right)^2+\left(x+3\right)^2=3\left(x-2\right)\left(x+1\right)\)
\(\Leftrightarrow2x^2-4x+2+x^2+6x+9=3x^2+3x-6x-6\)
\(\Leftrightarrow3x^2+2x+11=3x^2-3x-6\)
\(\Leftrightarrow5x+17=0\Leftrightarrow x=-\frac{17}{5}\)
b, \(\left(x+2\right)^2\left(x-3\right)=\left(x+1\right)^2\)
\(\Leftrightarrow x^3-3x^2+4x^2-12x+4x-12=x^2+2x+1\)
\(\Leftrightarrow x^3-8x-12=2x+1\)
\(\Leftrightarrow x^3-10x-13=0\)
\(\Leftrightarrow x\left(x^2-10\right)=13\)Lập bảng nhé, thú thật cái này phần này ko chắc:)
a) 2( x - 1 )2 + ( x + 3 )2 = 3( x - 2 )( x + 1 )
<=> 2( x2 - 2x + 1 ) + x2 + 6x + 9 = 3( x2 - x - 2 )
<=> 2x2 - 4x + 2 + x2 + 6x + 9 - 3x2 + 3x + 6 = 0
<=> 5x + 17 = 0
<=> 5x = -17
<=> x = -17/5
b) ( x + 2 )2( x - 3 ) = ( x + 1 )2
<=> ( x2 + 4x + 4 )( x - 3 ) = x2 + 2x + 1
<=> x3 + x2 - 8x - 12 - x2 - 2x - 1 = 0
<=> x3 - 10x - 13 = 0
Gồi đến đây chịu :) Chắc đề sai chỗ nào đấy
\(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)
ĐKXĐ : \(x\ne\pm2\)
\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)
\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+5x-14}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow12-\left(x^2+3x+2\right)+x^2+5x-14=0\)
\(\Leftrightarrow12-x^2-3x-2+x^2+5x-14=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)( không tmđk )
=> Phương trình vô nghiệm
\(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\left(đk:x\ne2;-2\right)\)
\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow12-\left(x^2+3x+3\right)+\left(x^2+5x-14\right)=0\)
\(\Leftrightarrow12-x^2+x^2-3x+5x-3-14=0\)
\(\Leftrightarrow2x-17+12=0\Leftrightarrow2x-5=0\Leftrightarrow x=\frac{5}{2}\left(tmđk\right)\)
a)\(A=\frac{6}{-2^2-3}\)
Ta có: \(x^2\ge0\Rightarrow2x^2+3\ge3\forall x\Rightarrow-2x^2-3\le-3\)
\(\Rightarrow A\ge-2\Rightarrow MinA=-2\)khi x=0
b) Ta có: \(x^2+2x+6=\left(x+1\right)^2+5\ge5\Rightarrow-x-2x-6\le-5\)
\(\Rightarrow B\ge\frac{-1}{5}\Rightarrow MinB=\frac{-1}{5}\)khi x=-1
c) Ta có:\(10x-x^2+3=-\left(x^2-10x+25\right)+28\le28\)\(\Rightarrow C\ge\frac{7}{28}=\frac{1}{4}\)