Giai phuong trinh
a.\(\left(x+1\right)^3+\left(x-2\right)^3=\left(2x-1\right)^3\)
b.\(x^4=24x+32\)
c.\(\left(x-6\right)^4+\left(x-8\right)^4=16\)
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Câu 1: Rút gọn các biểu thức sau:1. left(x+y-zright)^2+left(y-zright)^2+2zleft(z-yright)2. left(3x+4right)^2+left(x-4right)^2+2left(3x+4right)left(x-4right)3.left(3+1right)left(3^2+1right)left(3^4+1right)left(3^8+1right)left(3^{16}+1right)left(3^{32}+1right)4. 2xleft(2x-1right)^2-3xleft(x+3right)left(x-3right)-4xleft(x+1right)5. left(3+1right)left(3^2+1right)left(3^4+1right)left(3^8+1right)left(3^{16}+1right)left(3^{32}+1right)Câu 2: Tìm x1. 4left(x+1right)^2+left(2x-1right)^2-8left(x-1right)lef...
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Câu 1: Rút gọn các biểu thức sau:
1. \(\left(x+y-z\right)^2+\left(y-z\right)^2+2z\left(z-y\right)\)
2. \(\left(3x+4\right)^2+\left(x-4\right)^2+2\left(3x+4\right)\left(x-4\right)\)
3.\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
4. \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)\)
5. \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
Câu 2: Tìm x
1. \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=1\)
2. \(\left(3x+1\right)^2+\left(5x-2\right)^2=34\left(x+2\right)\left(x-2\right)\)
3. \(\left(x+3\right)^2+\left(x-2\right)^2=2x^2\)
4. \(4x^2-9-x\left(2x-3\right)=0\)
5. \(4x^2-12x+9=0\)
Câu 3: Tìm GTNN
D = \(\left(2x-1\right)^2+\left(x+2\right)^2\)
Câu 4: Cho \(a^2+b^2+c^2=ab+bc+ac\) . Chứng minh rằng a=b=c
Giải các phương trình sau:
a \(x^4-x^2-56=0\)
b \(\left(x-2\right)^4+\left(x+2\right)^4=32\)
c \(\left(x+3\right)^4+\left(x+5\right)^4=16\)
d \(\left(6-x\right)^4+\left(8-x\right)^4=80\)
a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)
Vậy x = 8 hoặc x = -7
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a: Ta có: \(x^4-x^2-56=0\)
\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)
\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)
\(\Leftrightarrow x^2-8=0\)
hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)
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28 tháng 2 2022 lúc 15:04
Tìm x biết:
\(a,\left(x-\dfrac{3}{4}\right)+50\%=\dfrac{1}{6}\)
\(b,\dfrac{1}{2}x-\dfrac{5}{6}x=\dfrac{7}{2}\)
\(c,\left(4-x\right)\left(3x+5\right)=0\)
\(d,\dfrac{x}{16}=\dfrac{50}{32}\)
\(e,\left(2x-3\right)+\dfrac{3}{2}=-\dfrac{1}{4}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
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Giai phuong trinh
\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
\(c,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(d,\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=3-15\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)
KL :....
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\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\) ĐKXĐ : \(x\ne0;2\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=2-2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
KL ::
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\(c,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\) ĐKXĐ : \(x\ne\pm2\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x^2+2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x+x+2+x^2-2x-x+2=2x^2+4\)
\(\Leftrightarrow0x=0\)
KL : PT vô số nghiệm
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Giải các phương trình sau:
a) x^2+dfrac{2x}{x-1}8
b) dfrac{x^2+2x+1}{x^2+2x+2}+dfrac{x^2+2x+2}{x^2+2x+3}dfrac{7}{6}
c) dfrac{x+4}{x-1}+dfrac{x-4}{x+1}dfrac{x+8}{x-2}+dfrac{x-8}{x+2}+6
d) left(x^2+6x+8right)left(x^2+8x+15right)24
e) left(x^2+x-2right)left(x^2+9x+18right)28
f) 3left(-x^2+2x+3right)^4-26x^2left(-x^2+2x+3right)^2-9x^40
g) x^4+6x^3+11x^2+6x+10
h) left(x-3right)left(x-5right)left(x-6right)left(x-10right)-24x^20
i) left(x+2right)^4+left(x+8right)^4272
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Giải các phương trình sau:
a) \(x^2+\dfrac{2x}{x-1}=8\)
b) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
c) \(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)
d) \(\left(x^2+6x+8\right)\left(x^2+8x+15\right)=24\)
e) \(\left(x^2+x-2\right)\left(x^2+9x+18\right)=28\)
f) \(3\left(-x^2+2x+3\right)^4-26x^2\left(-x^2+2x+3\right)^2-9x^4=0\)
g) \(x^4+6x^3+11x^2+6x+1=0\)
h) \(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=0\)
i) \(\left(x+2\right)^4+\left(x+8\right)^4=272\)
giải hết đống này chắc @@ quá,để tối đi,giờ t đi làm mấy bài ngắn ngắn
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a) x2+\(\dfrac{2x}{x-1}\)=8(ĐKXĐ : x ≠ 1
⇔ x2(x-1)+2x =8⇔ x3 - x2 +2x - 8=0
⇔x3 - 23 -x2+2x =0⇔ (x-2)(x2 +x+1) -x(x-2)
⇔(x-2)(x2 +1)⇒x =2
x2 +1 =0⇒x2 -1⇒x ∈∅(loại)
vậy x =2
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Giải phương trình :
a) \(x^2+\dfrac{2x}{x-1}=8\)
ĐKXĐ : \(x-1\ne0\Rightarrow x\ne1\)
Ta có : \(x^2+\dfrac{2x}{x-1}=8\)
\(\Leftrightarrow\) \(\dfrac{x^2\left(x-1\right)}{x-1}+\dfrac{2x}{x-1}=\dfrac{8\left(x-1\right)}{x-1}\)
\(\Rightarrow x^2\left(x-1\right)+2x=8\left(x-1\right)\)
\(\Leftrightarrow x^3-x^2+2x=8x-8\)
\(\Leftrightarrow x^3-x^2+2x-8x=-8\)
\(\Leftrightarrow x^3-x^2-6x+8=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(6x-8\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-2\left(3x-4\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x-1\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x-1=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)
Đối chiếu với ĐKXĐ ta được \(x\in\left\{\sqrt{2};\dfrac{4}{3}\right\}\) thỏa mãn.
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Giai cac bpt sau
a,\(\left(x+1\right)\left(2x-2\right)-3>-5x-\left(2x+1\right)\left(3-x\right)\)
b,\(\left(x-3^{ }\right)^2+4\left(2-x\right)>\left(x+7\right)\)
a: \(\Leftrightarrow2x^2-2-3>-5x+\left(2x+1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2-5>-5x+2x^2-6x+x-3\)
\(\Leftrightarrow2x^2-5>2x^2-10x-3\)
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: \(\Leftrightarrow x^2-6x+9+8-4x>x+7\)
\(\Leftrightarrow x^2-10x+17-x-7>0\)
\(\Leftrightarrow x^2-11x+10>0\)
=>x>10 hoặc x<1
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a: ⇔2x2−2−3>−5x+(2x+1)(x−3)⇔2x2−2−3>−5x+(2x+1)(x−3)
⇔2x2−5>−5x+2x2−6x+x−3⇔2x2−5>−5x+2x2−6x+x−3
⇔2x2−5>2x2−10x−3⇔2x2−5>2x2−10x−3
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: ⇔x2−6x+9+8−4x>x+7⇔x2−6x+9+8−4x>x+7
⇔x2−10x+17−x−7>0⇔x2−10x+17−x−7>0
⇔x2−11x+10>0⇔x2−11x+10>0
=>x>10 hoặc x<1
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Giải các phương trình sau:
a \(\left(x+2\right)\left(x+\text{4}\right)\left(x+6\right)\left(x+8\right)+16=0\)
b \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
c \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=0\)
d \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
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giải các phương trình sau
a. \(\left(x-3\right)\cdot\left(x-5\right)\cdot\left(x-6\right)\cdot\left(x-10\right)=24x^2\)
b. \(\left(x-6\right)^4+\left(x-8\right)^4=272\)
c. \(x^4-3x^3+2x^2-9x+9=0\)
Tìm x biết :a) left(x-2right)^3+6left(x+1right)^2-x^3+120b) left(x-5right)left(x+5right)-left(x+3right)^3+3left(x-2right)^2left(x+1right)^2-left(x+4right)left(x-4right)+3x^2c) left(2x+3right)^2+left(x-1right)left(x+1right)5left(x+2right)^2-left(x-5right)left(x+1right)+left(x+4right)^2d) left(1-3xright)^2-left(x-2right)left(9x+1right)left(3x-4right)left(3x+4right)-9left(x+3right)^2
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Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
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a) (x-2)3+6(x+1)2-x3+12=0
⇒ x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
⇒ x3-6x2+12x-8+6x2+12x+6-x3+12=0
⇒ 24x+10=0
⇒ 24x=-10
⇒ x=-5/12
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a.
PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)
\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)
b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.
c.
PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$
$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý)
Do đó pt vô nghiệm.
d.
PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$
$\Leftrightarrow 11x+3=-54x-97$
$\Leftrightarrow 65x=-100$
$\Leftrightarrow x=\frac{-20}{13}$
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