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Nếu ol thì tham khảo nah nguoiemtinhthong.
1.1
2x2+5x−1=7x3−1−−−−−√2x2+5x−1=7x3−1
⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)−−−−−−−−−−−−−−−√(1)⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)(1)
Đặt a=x−1−−−−−√;b=x2+x+1−−−−−−−−√;a≥0;b>0a=x−1;b=x2+x+1;a≥0;b>0
pt (1) trở thành 3a2+2b2−7ab=03a2+2b2−7ab=0
a=2ba=2b v a=13ba=13b
Các bạn tự giải quyết tiếp nhé.
1.2
TXĐ D=[1;+∞)D=[1;+∞)
đặt a=x−1−−−−−√4;b=x+1−−−−−√4;a,b≥0a=x−14;b=x+14;a,b≥0
pt (2) trở thành 3a2+2b2−5ab=03a2+2b2−5ab=0
⇔a=b⇔a=b v a=23ba=23b
...
1.3
D=[3;+∞)D=[3;+∞)
Đặt a=x2+4x−5−−−−−−−−−√;b=x−3−−−−−√;a,b≥0a=x2+4x−5;b=x−3;a,b≥0
pt (3) trở thành 3a+b=11a2−19b2−−−−−−−−−√3a+b=11a2−19b2
⇔2a2−6ab−20b2=0⇔2a2−6ab−20b2=0
⇒a=5b⇒a=5b
...
1.4
ĐK
⇔2x2−2x+2=3(x−2)x(x+1)−−−−−−−−−−−−√2x2−2x+2=3(x−2)x(x+1)
⇔(x2−2x)+2(x+1)=3(x2−2x)(x+1)−−−−−−−−−−−−−√2(x2−2x)+2(x+1)=3(x2−2x)(x+1)
Đặt x2−2x−−−−−−√=ax2−2x=a; x+1−−−−−√=bx+1=b (a;b\geq0)
⇔2a2+2b2=3ab
1.5
Đặt 4x2−4x−10=t4x2−4x−10=t (t \geq 0)
⇔t=t+4x2−2x−−−−−−−−−−√t=t+4x2−2x
⇔t2−t−4x2+2x=0t2−t−4x2+2x=0
Δ=1−4(2x−4x2)=(4x−1)2Δ=1−4(2x−4x2)=(4x−1)2
⇒t=1−2xt=1−2x hoặc t=2xt=2x
1.1
2.2+5.-1=7.3-1-----v2.2+5.-1=7.3-1
2(.2+x+1)+3(x-1)
3a+b=11a2-19b2
tóm tắt
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
Để olm giúp em em nhé!
a, \(\dfrac{x+2}{7x+42}\) = \(\dfrac{x+2}{7.\left(x+6\right)}\) = \(\dfrac{\left(x+2\right)\left(x-6\right)}{7\left(x-6\right)\left(x+6\right)}\) (đk \(x\ne\) \(\mp\) 6)
\(\dfrac{-13x}{x^2-36}\) = \(\dfrac{-13x}{\left(x-6\right)\left(x+6\right)}\) = \(\dfrac{-7.13.x}{7.\left(x-6\right).\left(x+6\right)}\) = \(\dfrac{-91x}{7.\left(x-6\right)\left(x+6\right)}\)
b, \(\dfrac{7}{4x+16}\) = \(\dfrac{7\left(x-4\right)}{4.\left(x+4\right).\left(x-4\right)}\) (đk \(x\ne\) \(\pm\) 4)
\(\dfrac{15}{x^2-16}\) = \(\dfrac{15.4}{\left(x-4\right)\left(x+4\right).4}\) = \(\dfrac{60}{4.\left(x-4\right).\left(x+4\right)}\)
3x.(x-2)-x2+2x=0
⇔3x2-6x-x2+2x=0
⇔2x2-4x=0
⇔2x(x-2)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
vậy x=0 và x=2
3x(x-2)-x^2+2x=0
<=>3x(x-2)-x(x-2)=0
<=>(3x-x)(x-2)=0
<=>2x(x-2)=0
<=>2x=0 hoặc x-2=0
<=>x=0 hoặc x=2
P=\(x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x=0\)
Vậy giá trị của P không phụ thuộc vào biến
Q=\(x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6=-8\)
Vậy giá trị của Q không phụ thuộc vào biến
Tick nha bạn 😘
\(a,=-15x^3+10x^4+20x^2\\ b,=2x^3+2x^2+4x-x^2-x-2=2x^3+x^2+3x-2\)
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)