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a)B = \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{7x+3}{9-x^2}\left(ĐK:x\ne\pm3\right)\)
= \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{7x+3}{x^2-9}\)
= \(\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-7x-3}{\left(x+3\right)\left(x-3\right)}\)
= \(\dfrac{3x^2-9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x+3}\)
b) \(\left|2x+1\right|=7< =>\left[{}\begin{matrix}2x+1=7< =>x=3\left(L\right)\\2x+1=-7< =>x=-4\left(C\right)\end{matrix}\right.\)
Thay x = -4 vào B, ta có:
B = \(\dfrac{-4.3}{-4+3}=12\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3x}{x+3}=\dfrac{-3}{5}< =>\dfrac{3x}{x+3}+\dfrac{3}{5}=0\)
<=> \(\dfrac{15x+3x+9}{5\left(x+3\right)}=0< =>x=\dfrac{-1}{2}\left(TM\right)\)
d) Để B nguyên <=> \(\dfrac{3x}{x+3}\) nguyên
<=> \(3-\dfrac{9}{x+3}\) nguyên <=> \(9⋮x+3\)
| x+3 | -9 | -3 | -1 | 1 | 3 | 9 |
| x | -12(C) | -6(C) | -4(C) | -2(C) | 0(C) | 6(C) |
a) A = \(\dfrac{1}{x-1}-\dfrac{4}{x+1}+\dfrac{8x}{\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x+1-4x+4+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x+5}{\left(x-1\right)\left(x+1\right)}=\dfrac{5}{x-1}\) => đpcm
b) \(\left|x-2\right|=3=>\left[{}\begin{matrix}x-2=3< =>x=5\left(C\right)\\x-2=-3< =>x=-1\left(L\right)\end{matrix}\right.\)
Thay x = 5 vào A, ta có:
A = \(\dfrac{5}{5-1}=\dfrac{5}{4}\)
c) Để A nguyên <=> \(5⋮x-1\)
| x-1 | -5 | -1 | 1 | 5 |
| x | -4(C) | 0(C) | 2(C) | 6(C) |
\(B=\dfrac{x^5-5x^3+4x}{30}=\dfrac{x\left(x^4-5x^2+4\right)}{30}=\dfrac{x\left(x^2-1\right)\left(x^2-4\right)}{30}=\dfrac{x\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)}{30}=\dfrac{\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)}{30}\).
Xét x nguyên. Trong 5 số x - 2, x - 1, x, x + 1, x + 2 tồn tại 1 số chia hết cho 2, 1 số chia hết cho 3, 1 số chia hết cho 5.
Do đó (x - 2)(x - 1)x(x + 1)(x + 2) luôn nguyên với mọi x nguyên.
Mặt khác tồn tại 2 số trong 5 số x - 2, x - 1, x, x + 1, x + 2 chia hết cho 2 mà 30 chia hết cho 2 nhưng không chia hết cho 4 nên B chia hết cho 2.
Vậy B khác 17 với mọi x nguyên.
1/ ĐKXĐ : \(x\ne1\)
\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\Leftrightarrow x=\dfrac{7}{19}\left(tm\right)\)
Vậy...
b/ \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) ĐKXĐ : \(x\ne-1\)
\(\Leftrightarrow12-28x=1+x\)
\(\Leftrightarrow11=29x\Leftrightarrow x=\dfrac{11}{29}\) \(\left(tm\right)\)
Vậy....
c/ ĐKXĐ : \(x\ne0\)
\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2x^2-12=2x^2+3x\)
\(\Leftrightarrow3x=-12\Leftrightarrow x=-4\) \(\left(tm\right)\)
Vậy...
4/ ĐKXĐ : \(x\ne-\dfrac{2}{3}\)
\(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)
\(\Leftrightarrow6x^2+4x-3x-2=5\)
\(\Leftrightarrow6x^2+x-7=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)
Vậy....
5,6 Tương tự nhé !
1)ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow21x-9-2x+2=0\)
\(\Leftrightarrow19x-7=0\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\dfrac{7}{19}\)(nhận)
Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)
2) ĐKXĐ: \(x\ne-1\)
Ta có: \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=x+1\)
\(\Leftrightarrow12-28x-x-1=0\)
\(\Leftrightarrow-29x+11=0\)
\(\Leftrightarrow-29x=-11\)
\(\Leftrightarrow x=\dfrac{11}{29}\)
Vậy: \(S=\left\{\dfrac{11}{29}\right\}\)
3) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-12=2x^2+6x\)
\(\Leftrightarrow2x^2-12-2x^2-6x=0\)
\(\Leftrightarrow-6x-12=0\)
\(\Leftrightarrow-6x=12\)
\(\Leftrightarrow x=-2\)
Vậy: S={-2}
Lời giải +HD
Hạ cấp
\(\dfrac{10x-4+3x}{75}=\dfrac{14x-x+3}{10}-x+1\)
rút gọn tử
\(\dfrac{13x-4}{75}=\dfrac{13x+3}{10}-x+1\)
Tách số hạng chứa x chuyển về VT (có thể sử pp cân bằng tử)
\(\left(\dfrac{13}{75}-\dfrac{13}{10}+1\right)x=\left(\dfrac{4}{75}+\dfrac{3}{10}+1\right)\)
Sử lý hệ số
\(\left(\dfrac{13.2-13.15+150}{150}\right)x=\left(\dfrac{4.2+3.15+150}{150}\right)\)
rút gọn hệ số
\(\left[13\left(2-15\right)+150\right]x=203\)
\(-19x=203\Rightarrow x=\dfrac{-203}{19}\)
a)
ĐKXĐ: \(x\ne-4\)
Để A nguyên thì \(3x+21⋮x+4\)
\(\Leftrightarrow3x+12+9⋮x+4\)
mà \(3x+12⋮x+4\)
nên \(9⋮x+4\)
\(\Leftrightarrow x+4\inƯ\left(9\right)\)
\(\Leftrightarrow x+4\in\left\{1;-1;3;-3;9;-9\right\}\)
\(\Leftrightarrow x\in\left\{-3;-5;-1;-7;5;-13\right\}\)(nhận)
Vậy: Để A nguyên thì \(x\in\left\{-3;-5;-1;-7;5;-13\right\}\)
b) ĐKXĐ: \(x\ne\dfrac{1}{2}\)
Để B nguyên thì \(2x^3-7x^2+7x+5⋮2x-1\)
\(\Leftrightarrow2x^3-x^2-6x^2+3x+4x-2+7⋮2x-1\)
\(\Leftrightarrow x^2\left(2x-1\right)-3x\left(2x-1\right)+2\left(2x-1\right)+7⋮2x-1\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-3x+2\right)+7⋮2x-1\)
mà \(\left(2x-1\right)\left(x^2-3x+2\right)⋮2x-1\)
nên \(7⋮2x-1\)
\(\Leftrightarrow2x-1\inƯ\left(7\right)\)
\(\Leftrightarrow2x-1\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow2x\in\left\{2;0;8;-6\right\}\)
hay \(x\in\left\{1;0;4;-3\right\}\)(nhận)
Vậy: \(x\in\left\{1;0;4;-3\right\}\)
a: \(A=\dfrac{x^2+1}{x}+\dfrac{x^3-1}{x^2-x}+\dfrac{x^4-x^3+x-1}{x-x^3}\)
\(=\dfrac{x^2+1}{x}+\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}-\dfrac{x^3\left(x-1\right)+\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+1}{x}+\dfrac{x^2+x+1}{x}-\dfrac{\left(x-1\right)\left(x^3+1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+1+x^2+x+1}{x}-\dfrac{x^2-x+1}{x}\)
\(=\dfrac{2x^2+x+2-x^2+x-1}{x}=\dfrac{x^2+2x+1}{x}=\dfrac{\left(x+1\right)^2}{x}\)
b: \(x^2+x=12\)
=>\(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>\(\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)
Thay x=3 vào A, ta được:
\(A=\dfrac{\left(3+1\right)^2}{3}=\dfrac{16}{3}\)
Khi x=-4 thì \(A=\dfrac{\left(-4+1\right)^2}{-4}=\dfrac{9}{-4}=-\dfrac{9}{4}\)
c: \(A-4=\dfrac{\left(x+1\right)^2}{x}-4\)
\(=\dfrac{\left(x+1\right)^2-4x}{x}\)
\(=\dfrac{x^2+2x+1-4x}{x}=\dfrac{x^2-2x+1}{x}=\dfrac{\left(x-1\right)^2}{x}\)>0 với mọi x>0
=>A>4
a. \(x^2y^3.35xy=5.7x^3y^4\)
\(\Leftrightarrow35x^3y^4=35x^3y^4\Rightarrowđpcm\)
\(b.x^2\left(x+2\right).\left(x+2\right)=x\left(x+2\right)^2.x\)
\(\Leftrightarrow x^2\left(x+2\right)^2=x^2\left(x+2\right)^2\Rightarrowđpcm\)
\(c.\left(3-x\right)\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)\)
\(\Leftrightarrow\left(3-x\right)\left(3-x\right)\left(3+x\right)=\left(3+x\right)\left(3-x\right)^2\)
\(\Leftrightarrow\left(3-x\right)^2\left(3+x\right)=\left(3-x\right)^2\left(3+x\right)\)
\(\Rightarrowđpcm\)
\(d.5\left(x^3-4x\right)=\left(10-5x\right)\left(-x^2-2x\right)\)
\(\Leftrightarrow5x^3-20x=5x^3-20x\Rightarrowđpcm\)
P(x) = \(\dfrac{x^5}{5}\) + \(\dfrac{x^3}{3}\) + \(\dfrac{7x}{15}\)
P(x) = \(\dfrac{3x^5+5x^3+7x}{15}\)
xét tử số 3x5 + 5x3 + 7x
= 3x5 + 3x3 + 6x + x + 2x3
= 3x5 + 3x3 + 6x + x ( 1 + 2x2)
nếu x không chia hết cho 3 ta có
vì x2 : 3 dư 1 ⇔ 2x2 : 3 dư 2 ⇔ 1 + 2x2 ⋮ 3⇔ A ⋮ 3 (2)
nếu x ⋮ 3 ⇔ A ⋮ 3 (1)
kết hợp (1) và (2) ta có
A ⋮ 3 ∀ x ϵ Z
A = 3x5 + 5x3 + 7x
A = 5x3 + 5x + 3x5 + 2x⋮
A = 5x3 + 5x + x ( 3x4 + 2)
nếu x ⋮ 5 thì A ⋮ 5 (*)
nếu x không chia hết cho 5 ta có : x2 : 5 dư 1 hoặc 4 (**)
nếu x2 : 5 dư 1 ⇔ 3x4 : 5 dư 1 ⇔ 3x4: 5 dư 3⇔ 3x4 + 2 ⋮ 5
nếu x2 : 5 dư 4 ⇔ x4 : 5 dư 16 ⇔ x4 : 5 dư 1 ⇔ 3x4 + 2 ⋮ 5
kết hợp (*) và(**) ta có A⋮ 5 ( ∀ x ϵ Z); (4)
kết hợp (3) và (4) A ⋮ 15 ∀ x ϵ Z
⇔ P(x) = \(\dfrac{A}{15}\) ϵ Z (∀x ϵZ) {đpcm}