1, Tính (x - 2)(x - 5) bằng:
A. \(x^2\) + 10. B. \(x^2\) + 7x + 10. C. \(x^2\)- 7x + 10. D. \(x^2\)- 3x + 10.
2, Kết quả của phép chia (2\(x^3\)- 5\(x^2\) + 6x – 15) : (2x – 5) là:
A. x + 3. B. x – 3. C. \(x^2\) – 3. D. \(x^2\) + 3 .
1.Tính
a, 5x^3yz . (-7x^2y^3)
b, 6x(x-5) -x(6x+3)
c, (x-9)(x^2-2x-1)
2.Cho A (x)=10-2x+4x^3-5x^2
B(x)=-10x^3-5x+6x^2-20
Tính A(x)+B(x); A(x)-B(x)
3.Tìm nghiệm
a,M(x)= 5x+20
b,N(x)=100x^2-49
c,P(x)=3x-15
b. 6x(x - 5) - x(6x + 3)
= x(6x - 30) - x(6x + 3)
= x(6x - 30 - 6x - 3)
= x(-33)
= -33x
1.Tính
a, 5x^3yz . (-7x^2y^3)
b, 6x(x-5) -x(6x+3)
c, (x-9)(x^2-2x-1)
2.Cho A (x)=10-2x+4x^3-5x^2
B(x)=-10x^3-5x+6x^2-20
Tính A(x)+B(x); A(x)-B(x)
3.Tìm nghiệm
a,M(x)= 5x+20
b,N(x)=100x^2-49
c,P(x)=3x-15
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
1.Tính
a, 5x^3yz . (-7x^2y^3)
b, 6x(x-5) -x(6x+3)
c, (x-9)(x^2-2x-1)
2.Cho A (x)=10-2x+4x^3-5x^2
B(x)=-10x^3-5x+6x^2-20
Tính A(x)+B(x); A(x)-B(x)
3.Tìm nghiệm
a,M(x)= 5x+20
b,N(x)=100x^2-49
c,P(x)=3x-15
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-x^2+3
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
bài 1:
a)7x(x2-7x+3) b)(x+6)(x-7)
c)(x-8)2 d) (3x+2)2
e)(x-4)(x+4)-(5-x)2
Bài 2:
a)2(x-7)-9=10 b)(2x-5)2-x(4x-3)=2x+50
\(1,\\ a,=7x^3-49x^2+21x\\ b,=x^2-x-42\\ c,=x^2-16x+64\\ d,=9x^2+12x+4\\ e,=x^2-16-25+10x-x^2=10x-41\\ 2,\\ a,\Rightarrow2\left(x-7\right)=19\\ \Rightarrow x-7=\dfrac{19}{2}\Rightarrow x=\dfrac{33}{2}\\ b,\Rightarrow4x^2-20x+25-4x^2+3x-2x=50\\ \Rightarrow-19x=25\Rightarrow x=-\dfrac{25}{19}\)
Gi ải các phương trình sau
a)x-3(2x-6)=21-(5x+3)
b)(x-2)(x+2)-(x-1)2=2(x+1)
c)\(\dfrac{9x+4}{6}\)=1-\(\dfrac{3x-5}{9}\)
d)\(\dfrac{6x+1}{x^2-7x+10}\)+\(\dfrac{5}{x-2}\)=\(\dfrac{3}{x-5}\)
a: \(x-3\left(2x-6\right)=21-\left(5x+3\right)\)
=>\(x-6x+18=21-5x-3\)
=>18=18(luôn đúng)
=>\(x\in R\)
b: \(\left(x-2\right)\left(x+2\right)-\left(x-1\right)^2=2\left(x+1\right)\)
=>\(x^2-4-x^2+2x-1=2x+2\)
=>2x-5=2x+2
=>-7=0(vô lý)
=>\(x\in\varnothing\)
c: \(\dfrac{9x+4}{6}=1-\dfrac{3x-5}{9}\)
=>\(\dfrac{3\left(9x+4\right)}{18}=\dfrac{18}{18}-\dfrac{2\left(3x-5\right)}{18}\)
=>3(9x+4)=18-2(3x-5)
=>27x+12=18-6x+10
=>27x+12=-6x+28
=>33x=16
=>\(x=\dfrac{16}{33}\left(nhận\right)\)
d: ĐKXĐ: \(x\notin\left\{2;5\right\}\)
\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(\dfrac{6x+1}{\left(x-2\right)\left(x-5\right)}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(6x+1+5\left(x-5\right)=3\left(x-2\right)\)
=>6x+1+5x-25=3x-6
=>11x-24=3x-6
=>8x=18
=>\(x=\dfrac{9}{4}\left(nhận\right)\)
a: x−3(2x−6)=21−(5x+3)
=>x−6x+18=21−5x−3
=>18=18(luôn đúng)
=>x∈R
b: (x−2)(x+2)−(x−1)2=2(x+1)
=>x2−4−x2+2x−1=2x+2
=>2x-5=2x+2
=>-7=0(vô lý)
=>x∈∅
c: 3(9x+4)18=1818−2(3x−5)18
=>3(9x+4)=18-2(3x-5)
=>27x+12=18-6x+10
=>27x+12=-6x+28
=>33x=16
=>6x+1x2−7x+10+5x−2=3x−5
=>x=94(nhận)
Quy đồng mẫu các phân thức sau
a) 10 =10
X+2 x+2
5 =5
2x-4 2(x-2)
1 =1
6-3x 3(x-2)
b) 7x-1 = 7x-1
2x2+6x 2x(x+3)
5-3x= 5-3x
X2-9 (x-3)(x+3)
c) X2+1=
X4 =
X2-1
d) 1 =
X2+3x-10
X =
X2+7x+10
Giải chi tiết hộ mik với ạ
a, [(x-2)(x+10)]/3-[(x+4)(x+10)]/12=[(x-2)(x+4)] b, (x+2)2/8-2(2x+1)=25+(x-2)2/3 c, (7x2-14x-5)/15=(2x+1)2/5-(x-1)2/3. d, [(7x+1)(x-2)]/10+2/5=(x-2)2/5+[(x-1)(x-3)]/2
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)