(2x+1)^2(2x-3)(2x+3)=14
Thực hiện phép tính:
a.(2x^2 + 2x - 4) : (x + 2)
b. ( 2x^3 - 5x^2 - x +1 ) : (2x + 1 )
c. (x^3 - 2x + 4)/(x + 2)
d. (x^3- 3x^2): (x - 3)
e. (x^4- x- 14): (x- 2)
f. (x^4- x- 14): (x- 2)
g. (17x^2- 6x^4+ 5x^3- 23x+ 7): (7- 3x^2- 2x)
giúp mik đi ạ mik đang cần gấp T-T
a: \(=\dfrac{2\left(x+2\right)\left(x-1\right)}{x+2}=2x-2\)
b: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=x^2-3x+1\)
c: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)
d: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)
`@` `\text {Ans}`
`\downarrow`
tìm x
17x – ( -16x – 37) = 2x +
-2x –3. (x – 17) = 34 – 2(-x + 25
17x + 3. ( -16x – 37) = 2x + 43 - 4x
103 -57: [-2. (2x – 1)2 – (-9)0] = -106
3x – 32 > -5x + 1
15 + 4x < 2x – 145
-3. (2x + 5) -16 < -4. (3 – 2x)
-2x + 15 < 3x – 7 < 19 – x
x + (x+1) + (x+2) + (x+3) + .... + 13 + 14 = 14
25 + 24 + 23 +...+ x + (x - 2) + (x – 3) = 25
17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)
lên mạng xem ik
hỏi google là đc hết mak
giải pt
a) \(\sqrt{x+2\sqrt{x-1}}+3\sqrt{x+8-6\sqrt{x-1}}=1-x\)
b) \(\sqrt{x\sqrt{x-1}-2x+2}+\sqrt{\left(x+3\right)\sqrt{x-1}-4x+4}=\sqrt{x-1}\)
c) \(\sqrt{14x+14\sqrt{14x-49}}+\sqrt{14x-14\sqrt{14x-49}}=14\)
d) \(\sqrt{2x-2\sqrt{2x-1}}-2\sqrt{2x+3-4\sqrt{2x-1}}+3\sqrt{2x+8-6\sqrt{2x-1}}=4\)
a/ ĐKXĐ: \(x\ge1\)
Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm
b/ \(x\ge1\)
\(\sqrt{\sqrt{x-1}\left(x-2\sqrt{x-1}\right)}+\sqrt{\sqrt{x-1}\left(x+3-4\sqrt{x-1}\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-1\right)^2}+\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-2\right)^2}=\sqrt{x-1}\)
Đặt \(\sqrt{x-1}=a\ge0\) ta được:
\(\sqrt{a\left(a-1\right)^2}+\sqrt{a\left(a-2\right)^2}=a\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\Rightarrow x=1\\\sqrt{\left(a-1\right)^2}+\sqrt{\left(a-2\right)^2}=\sqrt{a}\left(1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)
- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)
- Với \(0\le a\le1\) ta được:
\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)
- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)
c/ ĐKXĐ: \(x\ge\frac{49}{14}\)
\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)
\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)
Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)
Nên dấu "=" xảy ra khi và chỉ khi:
\(7-\sqrt{14x-49}\ge0\)
\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)
Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)
d/ ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}-2\sqrt{\left(\sqrt{2x-1}-2\right)^2}+3\sqrt{\left(\sqrt{2x-1}-3\right)^2}=4\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|-2\left|\sqrt{2x-1}-2\right|+3\left|\sqrt{2x-1}-3\right|=4\)
TH1: \(\sqrt{2x-1}\ge3\Rightarrow x\ge5\)
\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\sqrt{2x-1}-9=4\)
\(\Leftrightarrow\sqrt{2x-1}=5\)
\(\Leftrightarrow x=13\)
TH2: \(2\le\sqrt{2x-1}< 3\Rightarrow\frac{5}{2}\le x< 5\)
\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\left(3-\sqrt{2x-1}\right)=4\)
\(\Leftrightarrow\sqrt{2x-1}=2\Rightarrow x=\frac{5}{2}\)
TH3: \(1\le\sqrt{2x-1}< 2\Rightarrow1\le x< \frac{5}{2}\)
\(\sqrt{2x-1}-1-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)
\(\Leftrightarrow4=4\) (luôn đúng)
TH4: \(\frac{1}{2}\le x< 1\)
\(1-\sqrt{2x-1}-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)
\(\Leftrightarrow\sqrt{2x-1}=1\Rightarrow x=1\left(l\right)\)
Vậy nghiệm của pt là: \(\left[{}\begin{matrix}1\le x\le\frac{5}{2}\\x=13\end{matrix}\right.\)
Tìm x
1) (2x-1)(x+3)(2-x)=0
2)x^3 + x^2 + x + 1 = 0
3) 2x(x-3)+5(x-3) =0
4)x(2x-7)-(4x-14)=0
5) 2x^3 + 3x^2 + 2x + 3 = 0
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
Bài:Chia 1 biến đã sắp xếp 1)(2x^3+11x^2+18x-3):(2x+3) 2)(2x^3+11x^2+18x-3):(3x+3) 3)(2x^3+9x^2+5x+41):(2x^2-x+9) 4)(13x+41x^2+35x^3-14):(5x-2) 5)(5x^2-3x^3+15-9x):(5-3x) 6)(-4x^2+x^3-20+5x):(x-4)
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
7-(2x-1/3)^2=3
(2x+1/3)^2-3/8=1/8
12:[29-(x-2/3)^2]=3
(3x-1/2)^3+8/3=29/9-14/27
2(2x-1/3)^2+4/3=5/6+13/18
Tìm X
Tìm x biết:
a) 3x-|2x+1|=2
b)2.|5x-3|-2x=14
c)|x+1|+|x+2|+|x+3|=4x
d)|x-2|+|3-2x|=2x+1
e)|x-3|=(-2).|x+4|
A, x/2x+6 - x/2x+2= 3x+2/(x+1)(x+3)
B, 5/x+7+8/2x+14= 2/3
C, x-1/x-1/x+1= 2x-1/x^2+x
a, \(\frac{x}{2x+6}+\frac{x}{2x-2}=\frac{3x+2}{\left(x+1\right)\left(x+3\right)}\) Đkxđ : \(x\ne-1;x\ne-3\)
⇌ x(x + 1) - x(x - 3) = 2(3x + 2)
⇌ x2 + x - x2 - 3x = 6x + 4
⇌ -8x = 4
⇌ x = \(-\frac{1}{2}\) ( tm đk)
→ S = \(\left\{-\frac{1}{2}\right\}\)
b, \(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{2}{3}\) Đkxđ : \(x\ne-7\)
⇌ 30 + 24 = 2(x + 7)
⇌ 2x = 40
⇌ x = 20 (tmđk)
→ S = \(\left\{20\right\}\)
c, \(\frac{x-1}{\frac{x-1}{x+1}}=\frac{2x-1}{x^2+x}\) Đkxđ : \(x\ne-1\)
⇌ x = 2x - 1
⇌ x = 1 (tmđk)
→ S = \(\left\{1\right\}\)