Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Nguyễn TrươngNguyễn Việt LâmNguyenTruong Viet TruongKhôi BùiAkai HarumaÁnh LêDƯƠNG PHAN KHÁNH DƯƠNGPhùng Tuệ Minhsaint suppapong udomkaewkanjana
b) \(\left(x-7\right)\left(x-4\right)\left(x-5\right)\left(x-2\right)=72\)
\(\Leftrightarrow\) \(\left[\left(x-7\right)\left(x-2\right)\right].\left[\left(x-4\right)\left(x-5\right)\right]\) \(=72\)
\(\Leftrightarrow\) (\(x^2-9x+14\))(\(x^2-9x+20\)) \(=72\) (1)
Đặt \(x^2-9x+17=y\) .Khi đó (1) trở thành:
\(\left(y-3\right)\left(y+3\right)=72\)
\(\Leftrightarrow\) \(y^2-9=72\)
\(\Leftrightarrow\) \(y^2=81\) \(\Leftrightarrow\) \(y\) ∈ \(\left\{9;-9\right\}\)
+)Nếu \(y=9\) \(\Rightarrow\) \(x^2-9x+17=9\)
\(\Leftrightarrow\) \(x^2-9x-8=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\frac{9+\sqrt{113}}{2}\\x=\frac{9-\sqrt{113}}{2}\end{matrix}\right.\)
+)Nếu \(y=-9\) \(\Rightarrow x^2-9x+17=-9\)
\(\Leftrightarrow\) \(x^2-9x+26=0\)
\(\Leftrightarrow\)( \(x^2-2.x.\frac{9}{2}+\left(\frac{9}{2}\right)^2\)) \(+\frac{23}{4}\)\(=0\)
\(\Leftrightarrow\) \(\left(x-\frac{9}{2}\right)^2\)\(=-\frac{23}{4}\)( Vô lí,vì \(\left(x-\frac{9}{2}\right)^2\) ≥0)
Vậy phương trình có tập nghiệm S=\(\left\{\left(\frac{9+\sqrt{113}}{2}\right);\left(\frac{9-\sqrt{113}}{2}\right)\right\}\)
Bài 1 :
a)\(\frac{90}{x}-\frac{36}{x-6}=2\left(x\ne0,x\ne6\right)\)
\(\Leftrightarrow\frac{90x-540-36x-2x^2+12x}{x\left(x-6\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=18\left(tm\right)\\x=15\left(tm\right)\end{matrix}\right.\)
Kl:
b)\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\left(x\ne0,x\ne-10\right)\)
\(\Leftrightarrow\frac{12x+120+12x-x^2-10x}{12x\left(x+10\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=20\left(tm\right)\end{matrix}\right.\)
Kl:
c)\(\frac{x+7}{x-2}=\frac{x-2}{x+7}\left(x\ne2,x\ne-7\right)\)
\(\Leftrightarrow\left(x+7\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow10x+45=0\Leftrightarrow x=\frac{9}{2}\left(tm\right)\)
d)\(\frac{2x+5}{x+3}+\frac{3x+2}{x}=5\left(x\ne-3,x\ne0\right)\)
\(\Leftrightarrow2x^2+5x+3x^2+11x+6-5x^2-15=0\)
\(\Leftrightarrow x=\frac{9}{16}\left(tm\right)\)
Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)
Lời giải:
\(P(x)=x(x+2)(x+3)(x+5)-7\)
\(=[x(x+5)][(x+2)(x+3)]-7\)
\(=(x^2+5x)(x^2+5x+6)-7\)
\(=a(a+6)-7\) (đặt \(x^2+5x=a\) )
\(=a^2+6a-7=a^2-a+7a-7\)
\(=a(a-1)+7(a-1)=(a-1)(a+7)\)
\(=(x^2+5x-1)(x^2+5x+7)\)
-----------------
\(Q(x)=(4x-2)(10x+4)(5x+7)(2x+1)+17\)
\(=4(2x-1)(5x+2)(5x+7)(2x+1)+17\)
\(=4[(2x-1)(5x+7)][(5x+2)(2x+1)]+17\)
\(=4(10x^2+9x-7)(10x^2+9x+2)+17\)
\(=4a(a+9)+17\) (đặt \(10x^2+9x-7=a\)
\(=4a^2+36a+17=(2a+9)^2-8^2\)
\(=(2a+9-8)(2a+9+8)=(2a+1)(2a+17)\)
\(=(20x^2+18x-13)(20x^2+18x+3)\)
\(R(x)=(3x+2)(3x-5)(x-1)(9x+10)+24x^2\)
\(=[(3x+2)(3x-5)][(x-1)(9x+10)]+24x^2\)
\(=(9x^2-9x-10)(9x^2+x-10)+24x^2\)
\(=(a-9x)(a+x)+24x^2\) (đặt \(9x^2-10=a\) )
\(=a^2-8ax+15x^2=(a^2-5ax)-(3ax-15x^2)\)
\(=a(a-5x)-3x(a-5x)=(a-3x)(a-5x)\)
\(=(9x^2-3x-10)(9x^2-5x-10)\)
--------------------------
\(H(x)=(x-18)(x-7)(x+35)(x+90)-67x^2\)
\(=[(x-18)(x+35)][(x-7)(x+90)]-67x^2\)
\(=(x^2+17x-630)(x^2+83x-630)-67x^2\)
\(=a(a+66x)-67x^2\) (đặt \(x^2+17x-630=a\) )
\(=a^2-ax+67ax-67x^2\)
\(=a(a-x)+67x(a-x)=(a-x)(a+67x)\)
\(=(x^2+16x-630)(x^2+84x-630)\)
a.(x-2)(x+2)(x2-10)=72
⇔(x2-4)(x2-10)=72
đặt x2-7=t.Ta có pt ẩn t:
(t2+3)(t2-3)=72
⇔t2-9-72=0
⇔t2-81=0
⇔(t-9)(t+9)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-7-9=0\\x^2-7+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=16\\x^2=-2\left(loại\right)\end{matrix}\right.\)
⇔x2=16
⇔x=4 hoặc x=-4
vậy pt đã cho có tập nghiệm là S=\(\left\{4;-4\right\}\)
b: =>(x^2+6x+5)(x^2+6x+8)=40
=>(x^2+6x)^2+13(x^2+6x)=0
=>(x^2+6x)(x^2+6x+13)=0
=>x=0 hoặc x=-6
c: \(\left(2x^2+3x-1\right)^2-5\left(2x^2+3x+3\right)+24=0\)
Đặt 2x^2+3x-1=a
=>a^2-5(a+4)+24=0
=>a^2-5a-20+24=0
=>(a-1)(a-4)=0
=>(2x^2+3x-2)(2x^2+3x-5)=0
=>\(x\in\left\{\dfrac{1}{2};-2;1;-\dfrac{5}{2}\right\}\)
b; x(x-1)(x+1)(x+2)-24
=(x2+x)(x2+x-2)-24
Đặt x2+x=k khi đó k(k-2)-24=k2-2k-24
=(k2-2k+1)-25=(k-1)2-52
=(k-1-5)(k-1+5)=(k-6)(k+4)
c; (x+2)(x-2)(x2-10)-72
=(x2-4)(x2-10)-72
Đặt x2-7=k khi đó (k-3)(k+3)-72=k2-9-72
=k2-81=(k-9)(k+9)=(x2-7-9)(x2-7+9)
=(x2-16)(x2+2)=(x-4)(x+4)(x2+2)
d; (x-7)(x-5)(x-4)(x-2)-72
=(x2-9x+14)(x2-9x+8)-72
Đặt x2-9x+11=k khi đó (k+3)(k-3)-72=k2-9-72
=k2-81=(k-9)(k+9)=(x2-9x+11-9)(x2-9x+11+9)
=(x2-9x+2)(x2-9x+20)
=(x2-9x+2)(x2-4x-5x+20)
=(x2-9x+2)(x-4)(x-5)