Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
DKXD:x\(\ne\)\(\pm\)2
\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\Leftrightarrow\dfrac{\left(x-2\right)^2-3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x-22}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow\)x2-4x+4-3x-6=2x-22\(\Leftrightarrow\)x2-4x-3x-2x+22+4-6=0\(\Leftrightarrow\)x2-9x+20=0
\(\Leftrightarrow\)(x-4)(x-5)\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\Leftrightarrow x=4\left(TM\right)\\x-5=0\Leftrightarrow x=5\left(TM\right)\end{matrix}\right.\)
Vậy tập nghiệm của PT là:S={4;5}
Bài 17)
(x - 2)^4 + (x - 6)^4 = 82
Đặt t = x + 3
=> x + 2 = t - 1; x + 4 = t + 1.
ta có pt: (t - 1)^4 + (t + 1)^4 = 82
<=>[(t -1)²]² + [(t + 1)²]² = 82
<=> (t² - 2t + 1)² + (t² + 2t + 1)² = 82
<=> (t²+1)² - 4t(t²+1) + 4t² + (t²+1)² + 4t(t²+1) + 4t² = 82
<=> (t² + 1)² + 4t² = 41
<=> t^4 + 6t² + 1 = 41
<=> (t²)² + 6t² - 40 = 0
<=> t² = -10 (loại) hoặc t² = 4
<=> t = 2 hoặc t = -2
với t = -2 => x = -5
với t = 2 => x = -1
vậy pt có hai nghiệm là : x = -1 hoặc x = -5
Bài 18: Phương trình đã cho được viết thành: $${({x^2} + 6x + 10)^2} + (x + 3)\left[ {3\left( {{x^2} + 6x + 10} \right) + 2\left( {x + 3} \right)} \right] = 0$$
Đặt $u = {x^2} + 6x + 10 > 0,v = x + 3$, suy ra:
$${u^2} + v\left( {3u + 2v} \right) = 0 \Leftrightarrow \left( {u + v} \right)\left( {u + 2v} \right) = 0 \Leftrightarrow \left[ \begin{gathered}
u + v = 0 \\
u + 2v = 0 \\
\end{gathered} \right.$$
$$ \Leftrightarrow \left[ \begin{gathered}
{x^2} + 6x + 10 + x + 3 = 0 \\
{x^2} + 6x + 10 + 2\left( {x + 3} \right) = 0 \\
\end{gathered} \right. \Leftrightarrow \left[ \begin{gathered}
{x^2} + 7x + 13 = 0 \\
{x^2} + 8x + 16 = 0 \\
\end{gathered} \right. \Leftrightarrow x = - 4$$
Đặt \(\left(a-1\right)^2=t\)
Ta có: \(\left(a-1\right)^4-11\left(a-1\right)^2+30\)
\(=t^2-11t+30\)
\(=t\left(t-5\right)-6\left(t-5\right)=\left(t-5\right)\left(t-6\right)\)
\(=\left[\left(a-1\right)^2-5\right]\left[\left(a-1\right)^2-6\right]\)
\(=\left(a^2-2a-4\right)\left(a^2-2a-5\right)\)
Đặt \(a^2-2a=k\)
Ta có: \(3\left(a-1\right)^4-18\left(a^2-2a\right)-3\)
\(=3\left(a^2-2a+1\right)^2-18\left(a^2-2a\right)-3\)
\(=3\left(k+1\right)^2-18k-3\)
\(=3k^2+6k+3-18k-3\)
\(=3k^2-12k=3k\left(k-4\right)\)
\(=3\left(a^2-2a\right)\left(a^2-2a-4\right)\)(Ở đây bạn ghi thêm điều kiện nhé)
Khi đó: \(N=\frac{\left(a^2-2a-4\right)\left(a^2-2a-5\right)}{3\left(a^2-2a\right)\left(a^2-2a-4\right)}=\frac{a^2-2a-5}{3\left(a^2-2a\right)}\)
a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=5x\left(2-x\right)-11\left(x+2\right)\)
=>-x^2+2x-1=10x-5x^2-11x-22
=>-x^2+2x-1=-5x^2-x-22
=>4x^2+3x+21=0
=>PTVN
b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)
=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)
=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80
=>20x+16=32x-80
=>-12x=-96
=>x=8
c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)
=>6x-18+7x-35=13x+4
=>-53=4(loại)
d: =>3(2x-1)-5(x-2)=3(x+7)
=>6x-3-5x+10=3x+21
=>3x+21=x+7
=>x=-7
e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1
=>-9x^2+9x-9=-9x^2+1
=>9x=10
=>x=10/9
Lời giải:
\(N=\frac{(a-1)^4-11(a-1)^2+30}{3(a-1)^4-18(a^2-2a+1)+15}=\frac{(a-1)^4-11(a-1)^2+30}{3(a-1)^4-18(a-1)^2+15}\)
Đặt \((a-1)^2=t\Rightarrow N=\frac{t^2-11t+30}{3t^2-18t+15}\)
\(=\frac{t^2-11t+30}{3(t^2-6t+5)}=\frac{(t-5)(t-6)}{3(t-1)(t-5)}\)
\(=\frac{t-6}{3(t-1)}=\frac{(a-1)^2-6}{3(a-1)^2-3}\)
dễ thì giải cho người ta đi,bạn thông minh hơn thì thay vì ns người khác thì giúp người khác sẽ tốt hơn đó
a/ \(\dfrac{2x^2-20x+50}{3x+3}\cdot\dfrac{x^2-1}{4\left(x-5\right)^2}=\dfrac{2\left(x^2-10x+25\right)\cdot\left(x^2-1\right)}{3\left(x+1\right)\cdot4\left(x-5\right)^2}=\dfrac{2\left(x-5\right)^2\left(x-1\right)\left(x+1\right)}{12\left(x+1\right)\left(x-5\right)^2}=\dfrac{x+1}{6}\)
b/ \(\dfrac{6x-3}{5x^2+x}\cdot\dfrac{25x^2+10x+1}{1-8x^2}=-\dfrac{3\left(1-2x\right)\cdot\left(5x+1\right)^2}{x\left(5x+1\right)\left(1-2x\right)\left(1+2x+4x^2\right)}=\dfrac{3\left(5x+1\right)}{x\left(4x^2+2x+1\right)}\)
c/ \(\dfrac{3x^2-x}{x^2-1}\cdot\dfrac{1-x^4}{\left(1-3x\right)^3}=\dfrac{x-3x^2}{1-x^2}\cdot\dfrac{\left(1-x^2\right)\left(1+x^2\right)}{\left(1-3x\right)^3}=\dfrac{x\left(1-3x\right)\left(1-x^2\right)\left(1+x^2\right)}{\left(1-x^2\right)\left(1-3x\right)^3}=\dfrac{x\left(x^2+1\right)}{\left(1-3x\right)^3}\)
\(\dfrac{3\left(x-11\right)}{4}=\dfrac{3\left(x+1\right)}{5}\\ \Rightarrow\dfrac{3x-33}{4}=\dfrac{3x+3}{5}=\dfrac{3x-33-3x-3}{4-5}\\ =\dfrac{-36}{-1}=36\\ \Rightarrow3\left(x-11\right)=144\\ \Rightarrow x-11=48\\ \Rightarrow x=59\)
Đề như vậy thì chỉ cần 2 pth đầu là đc chắc bn ghi đề sai, mk lm theo đề như này nhé:
\(\dfrac{3\left(x-11\right)}{4}+\dfrac{3\left(x+1\right)}{5}=\dfrac{2\left(2x-5\right)}{10}\)
\(\Leftrightarrow\dfrac{15\left(x-11\right)}{20}+\dfrac{12\left(x+1\right)}{20}=\dfrac{4\left(2x-5\right)}{20}\)
\(\Leftrightarrow15\left(x-11\right)+12\left(x+1\right)=4\left(2x-5\right)\)
\(\Leftrightarrow15x-165+12x+12=8x-20\)
\(\Leftrightarrow15x+12x-8x=-20-12+165\)
\(\Leftrightarrow19x=133\Leftrightarrow x=7\)
Vậy x = 7
\(18+\dfrac{1}{11}\times\left(x-18\right)=36+\dfrac{1}{11}\times\left[\dfrac{10}{11}\times\left(x-18\right)-36\right]\)
\(\Leftrightarrow\dfrac{198}{11}+\dfrac{1}{11}\times\left(x-18\right)=36+\dfrac{1}{11}\times\left[\dfrac{10}{11}\times\left(x-18\right)-\dfrac{396}{11}\right]\)
\(\Leftrightarrow\dfrac{198+x-18}{11}=36+\dfrac{1}{11}\times\dfrac{10x-180-396}{11}\)
\(\Leftrightarrow\dfrac{180+x}{11}=36+\dfrac{10x-576}{121}\)
\(\Leftrightarrow\dfrac{1980+11x}{121}=\dfrac{4356}{121}+\dfrac{10x-576}{121}\)
\(\Leftrightarrow1980+11x=4356+10x-576\)
\(\Leftrightarrow11x-10x=4356-1980-576\)
\(\Leftrightarrow x=1800\)