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a: ĐKXĐ: x<>-3/2
Để \(\frac{5x+11}{2x+3}\) là số nguyên thì \(5x+11\vdots2x+3\)
=>\(10x+22\vdots2x+3\)
=>\(10x+15+7\vdots2x+3\)
=>7⋮2x+3
=>2x+3∈{1;-1;7;-7}
=>2x∈{-2;-4;4;-10}
=>x∈{-1;-2;2;-5}
b: ĐKXĐ: x<>1/3
Để \(\frac{5x-4}{3x-1}\) là số nguyên thì 5x-4⋮3x-1
=>15x-12⋮3x-1
=>15x-5-7⋮3x-1
=>-7⋮3x-1
=>3x-1∈{1;-1;7;-7}
=>3x∈{2;0;8;-6}
=>x∈\(\left\lbrace\frac23;0;\frac83;-2\right\rbrace\)
mà x nguyên
nên x∈{0;-2}
c: ĐKXĐ: x<>-2/3
Để \(\frac{5x}{3x+2}\) là số nguyên thì 5x⋮3x+2
=>15x⋮3x+2
=>15x+10-10⋮3x+2
=>-10⋮3x+2
=>3x+2∈{1;-1;2;-2;5;-5;10;-10}
=>3x∈{-1;-3;0;-4;3;-7;8;-12}
=>x∈{-1/3;-1;0;-4/3;1;-7/3;8/3;-4}
mà x nguyên
nên x∈{-1;0;1;-4}
d:
ĐKXĐ: x<>-3/4
Để \(\frac{7x+7}{4x+3}\) là số nguyên thì 7x+7⋮4x+3
=>28x+28⋮4x+3
=>28x+21+7⋮4x+3
=>7⋮4x+3
=>4x+3∈{1;-1;7;-7}
=>4x∈{-2;-4;4;-10}
=>x∈\(\left\lbrace-\frac12;-1;1;-\frac52\right\rbrace\)
mà x nguyên
nên x∈{-1;1}
e: ĐKXĐ: x∈R
Để \(\frac{2x^2-x+2}{x^2-x+2}\) là số nguyên thì \(2x^2-x+2\vdots x^2-x+2\)
=>\(2x^2-2x+4+x-2\vdots x^2-x+2\)
=>\(x-2\vdots x^2-x+2\)
=>\(\left(x-2\right)\left(x+1\right)\vdots x^2-x+2\)
=>\(x^2-x-2\vdots x^2-x+2\)
=>\(x^2-x+2-4\vdots x^2-x+2\)
=>\(-4\vdots x^2-x+2\)
mà \(x^2-x+2=\left(x-\frac12\right)^2+\frac74\ge\frac74\forall x\)
nên \(x^2-x+2\in\left\lbrace2;4\right\rbrace\)
TH1: \(x^2-x+2=2\)
=>\(x^2-x=0\)
=>x(x-1)=0
=>\(\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)
Thay lại vào phân số, ta thấy x=0 thỏa mãn
TH2: \(x^2-x+2=4\)
=>\(x^2-x-2=0\)
=>(x-2)(x+1)=0
=>\(\left[\begin{array}{l}x=2\\ x=-1\end{array}\right.\)
Thay lại vào phân số, ta thấy x=2 thỏa mãn
Vậy: x∈{0;2}
\(\dfrac{x^3-x^2-x+1}{x^4-2x^2+1}=\dfrac{x^2\left(x-1\right)-\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}=\dfrac{1}{x+1}\)
\(\dfrac{5x^3+10x^2+5x}{x^3+3x^2+3x+1}=\dfrac{5x\left(x+1\right)^2}{\left(x+1\right)^3}=\dfrac{5x}{x+1}\)
Bài 2:
a: Ta có: \(x\left(2x-1\right)-2x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Tìm số tự nhiên x: \(2^{x-1}+5.2^{x-2}=224\Leftrightarrow2.2^{x-2}+5.2^{x-2}=224\)
\(\Leftrightarrow2^{x-2}.\left(5+2\right)=224\Leftrightarrow2^{x-2}.7=224\)
\(\Rightarrow2^{x-2}=32\Leftrightarrow2^{x-2}=2^5\)\(\Rightarrow x-2=5\Leftrightarrow x=7\)
Vậy x=7
Tìm x biết: \(\frac{3}{7}=\frac{2x+1}{3x+5}\)
\(\Rightarrow3\left(3x+5\right)=7\left(2x+1\right)\Leftrightarrow9x+15=14x+7\)
\(\Leftrightarrow14x+7-\left(9x+15\right)=0\Rightarrow5x+\left(-8\right)=0\)
\(\Leftrightarrow5x=8\Rightarrow x=\frac{8}{5}\)
Vậy x=8/5
ta có : \(A=\dfrac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\dfrac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\dfrac{\sqrt{x+3}}{\sqrt{x-3}}\)
ta có : \(B=\dfrac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}=\dfrac{\left(x+2\right)\left(x+3\right)+x\sqrt{ 9-x^2}}{x\left(3-x\right)+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\dfrac{\sqrt{x+3}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(x\sqrt{3-x}+\left(x+2\right)\sqrt{x+3}\right)}=\dfrac{\sqrt{x+3}}{\sqrt{3-x}}\)
Bài 3
a) 2x(x - 3) - x + 3 = 0
2x(x - 3) - (x - 3) = 0
(x - 3)(2x - 1) = 0
x - 3 = 0 hoặc 2x - 1 = 0
*) x - 3 = 0
x = 3
*) 2x - 1 = 0
2x = 1
x = 1/2
Vậy x = 1/2; x = 3
b) (3x - 1)(2x + 1) - (x + 1)² = 5x²
6x² + 3x - 2x - 1 - x² - 2x - 1 - 5x² = 0
(6x² - x² - 5x²) + (3x - 2x - 2x) = 0 + 1 + 1
-x = 2
x = -2
Bài 2
a) 5x² + 30y
= 5(x² + 6y)
b) x³ - 2x² - 4xy² + x
= x(x² - 2x - 4y² + 1)
= x[(x² - 2x + 1) - 4y²]
= x[(x - 1)² - (2y)²]
= x(x - 1 - 2y)(x - 1 + 2y)
1)
a) \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)\)
\(=\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]\)
\(=\left(x+2\right)\left(x+2-x+2\right)\)
\(=4\left(x+2\right)\)
b) \(x+2x^2+2x^3\)
\(=x\left(2x+2x^2+1\right)\)
1) a. \(\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)\)
b. \(x\left(1+2x+2x^2\right)\)
2) a. \(=x^2-4-\left(x^2+4x+3\right)=x^2-4-x^2-4x-3=-4x-7\)
b. Áp dụng dạng \(\left(a+b\right)^2=a^2+b^2+2ab\)
\(\left(2x+1\right)^2+\left(3x-1\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
Đề bài thiếu dữ kiện rồi bạn
Để biểu thức là số nguyên nha