Tìm x,biết:
2=\(\frac{1954.0,24+76.19,54}{977.\left(x-4\right)}\)
\(2=\frac{1954.0,24+76.19,54}{977.\left(x-4\right)}\)
\(2=\frac{1954\cdot0,24+76\cdot19,54}{977\cdot\left(x-4\right)}\)
Tìm x biết \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+4\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
Tìm x thuộc Q biết: \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
ĐKXĐ:\(x\ne\left\{-2;-4;-8;-14\right\}\)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
\(\Leftrightarrow2\left(x+8\right)\left(x+14\right)+4\left(x+2\right)\left(x+14\right)+6\left(x+2\right)\left(x+4\right)=x\left(x+8\right)\left(x+14\right)\)
\(\Leftrightarrow2x^2+44x+224+4x^2+64x+112+6x^2+36x+48=x^3+22x^2+112x\)
\(\Leftrightarrow12x^2+144x+384=x^3+22x^2+112x\)
\(\Leftrightarrow x^3+22x^2-12x^2+112x-144x-384=0\)
\(\Leftrightarrow x^3+10x^2-32x-384=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2+16x+64\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+8\right)^2=0\)
\(\Leftrightarrow x=6\)(x=-8 loại vì x=-8 thì PT không xác định)
Điều kiện: x+ 2 \(\ne\) 0 ; x+ 4 \(\ne\) 0; x+ 8 \(\ne\) 0 ; x + 14 \(\ne\) 0
<=> \(\frac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\frac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\frac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
<=> \(\frac{x+4}{\left(x+2\right)\left(x+4\right)}-\frac{x+2}{\left(x+2\right)\left(x+4\right)}+\frac{x+8}{\left(x+4\right)\left(x+8\right)}-\frac{x+4}{\left(x+4\right)\left(x+8\right)}+\frac{x+14}{\left(x+8\right)\left(x+14\right)}-\frac{x+8}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)<=> \(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
<=> \(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
<=> \(\frac{12\left(x+4\right)}{\left(x+2\right)\left(x+14\right)\left(x+4\right)}=\frac{x\left(x+14\right)}{\left(x+2\right)\left(x+4\right)\left(x+14\right)}\)
<=> 12(x + 4) = x (x + 14)
<=> 12x + 48 = x2 + 14 x
<=> x2 + 2x - 48 = 0
<=> x2 + 8x - 6x - 48 = 0
<=> x(x + 8) - 6 (x + 8) = 0
<=> (x - 6)(x + 8) = 0 <=> x - 6 = 0 (do x + 8 \(\ne\) 0)
<=> x = 6
Vậy x = 6
Tìm x biết: \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\)=\(\left(x+4\right)^2\)
Đặt \(t=\left(x+\frac{1}{x}\right)^2\)\(\Rightarrow\)\(x^2+\frac{1}{x^2}=t-2\)điều kiện t>=0,x # 0
Phương trình trở thành
8t +4(t-2)2 - 4(t-2)2t =(x+4)2
8t + 4t2 - 16t + 16 -4t3 + 16t2 - 16t=(x+4)2
-4t3 + 20t2 -24t=x2 +8x
-4t(t2 -5t +6)=x(x+8)
-4t(t-2)(t-3)=x(x+8)
Mình chỉ giúp dược tới đó
tìm x biết
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
tìm x biết :
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)
\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(4-x+x=x\)
\(\Leftrightarrow x=4\)
lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!
tìm x biết \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2...\)
ai làm nhanh và đúng thì đc 3 tick nha
đó chính là -4 minh khong muon giai ra ta lau lam ban
rút 4 ra ngoài nhan bạn 4(2(x+1/x)^2+(x^2+1/x^2)^2-(x^2+1/x^2)(x+1/x)^2=(x+4)^2
mik xét cái này cho dễ nhìn nhan
2(x+1/x)^2-(x^2+1/x^2)(x+1/x)^2
= (x+1/x)^2(2-x^2-1/x^2)
= -(x+1/x)^2(x^2-2+1/x^2)
= -(x+1/x)^2(x-1/x)^2=-(x^2-1/x^2)^2
thế ở trên ta có
4(-(x^2-1/x^2)^2+(x^2+1/x^2)^2)=(x+4)^2
4(-x^4+2-1/x^4+x^4+2+1/x^4)=x^2+8x+16
4.4=x^2+8x+16
suy ra x^2+8x=0
x(x+8)=0
suy ra x=0 hoặc x=-8
mak nhìn để bài thì x=0 ko được nên x=-8
Tìm số nguyên x biết \(\left(x^2-\frac{4}{25}\right)\left(x^2-4\right)\left(x^2-\frac{16}{9}\right)\left(x^2-10\right)< 10\)