Tính:
a) x2 - 4/8x . x +190/x-2 + x2 - 4/8x.x-194/x-2
b) 1/(x-5)(x-4) + 1/(x-4)(x-3) + 1/(x-3)(x-2) - 1/x-5
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Mọi người làm nhanh hộ e với ạ, T7 e nộp rBài 1.Tính:a. x2(x–2x3) b. (x2+ 1)(5–x) c. (x–2)(x2+ 3x–4) d. (x–2)(x–x2+ 4)e. (x2–1)(x2+ 2x) f. (2x–1)(3x + 2)(3–x) g. (x + 3)(x2+ 3x–5)h (xy–2).(x3–2x–6) i. (5x3–x2+ 2x–3).(4x2–x + 2)Bài 2.Tính:a. (x–2y)2 b. (2x2+3)2 c. (x–2)(x2+ 2x + 4) d. (2x–1)2Bài 3: Rút gọn biểu thứca.(6x + 1)2+ (6x–1)2–2(1 + 6x)(6x–1)b. x(2x2–3)–x2(5x + 1) + x2.c. 3x(x–2)–5x(1–x)–8(x2–3)Bài 4: Tìm x, biếta. (x–2)2–(x–3)(x + 3) 6.b. 4(x–3)2–(2x–1)(2x + 1) 10c. (x–4...
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Mọi người làm nhanh hộ e với ạ, T7 e nộp r
Bài 1.
Tính:
a. x2(x–2x3) b. (x2+ 1)(5–x) c. (x–2)(x2+ 3x–4) d. (x–2)(x–x2+ 4)
e. (x2–1)(x2+ 2x) f. (2x–1)(3x + 2)(3–x) g. (x + 3)(x2+ 3x–5)
h (xy–2).(x3–2x–6) i. (5x3–x2+ 2x–3).(4x2–x + 2)
Bài 2.
Tính:
a. (x–2y)2 b. (2x2+3)2 c. (x–2)(x2+ 2x + 4) d. (2x–1)2
Bài 3: Rút gọn biểu thức
a.(6x + 1)2+ (6x–1)2–2(1 + 6x)(6x–1)
b. x(2x2–3)–x2(5x + 1) + x2.
c. 3x(x–2)–5x(1–x)–8(x2–3)
Bài 4: Tìm x, biết
a. (x–2)2–(x–3)(x + 3) = 6.
b. 4(x–3)2–(2x–1)(2x + 1) = 10
c. (x–4)2–(x–2)(x + 2) = 6.
d. 9 (x + 1)2–(3x–2)(3x + 2) = 10
Bài 5:Phân tích các đa thức sau thành nhân tử
a. 1–2y + y2
b. (x + 1)2–25
c. 1–4x2
d. 8–27x3
e. 27 + 27x + 9x2+ x3
f. 8x3–12x2y +6xy2–y3
g. x3+ 8y3
Bài 6:Phân tích các đa thức sau thành nhân tử
a. 3x2–6x + 9x2
b. 10x(x–y)–6y(y–x)
c. 3x2+ 5y–3xy–5x
d. 3y2–3z2+ 3x2+ 6xy
e. 16x3+ 54y3
f. x2–25–2xy + y2
g. x5–3x4+ 3x3–x2
.
Bài 7: Phân tích đa thức thành nhân tử
a. 5x2–10xy + 5y2–20z2
b. 16x–5x2–3
c. x2–5x + 5y–y2
d. 3x2–6xy + 3y2–12z2
e. x2+ 4x + 3
f. (x2+ 1)2–4x2
g. x2–4x–5
Bài 5:
a. 1 - 2y + y2
= (1 - y)2
b. (x + 1)2 - 25
= (x + 1)2 - 52
= (x + 1 - 5)(x + 1 + 5)
= (x - 4)(x + 6)
c. 1 - 4x2
= 12 - (2x)2
= (1 - 2x)(1 + 2x)
d. 8 - 27x3
= 23 - (3x)3
= (2 - 3x)(4 + 6x + 9x2)
e. (đề hơi khó hiểu ''x3'' !?)
g. x3 + 8y3
= (x + 2y)(x2 - 2xy + y2)
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Thực hiện phép tính:
a,4.(x+3)/3x2-x : x2+3x/1-3x
b, x+1/x2-2x-8 . 4-x/x2+x
c, 9x+5/2(x-1)(x+3)2- 5x-7/2(x-1)(x+3)2
d, 18/(x-3)(x2-9)-3/x^2-6x+9-x/x^2-9
e, 1/x2-x+1+1/1-x2+2/x3+1
Bài 1. Thực hiện phép tính:
a) 2xy(x2+ xy - 3y2)
b) (x + 2)(3x2 - 4x)
c) (x3 + 3x2 - 8x - 20) : (x + 2)
d) (x + y)2 + (x – y)2 – 2(x + y)(x - y) e) (a + b)3 - (a – b)3 – 2b3
f) 2x2(x – 2)+ 3x(x2 – x – 2) –5(3 – x2)
g) (x – 1)(x – 3) – (4 – x)(2x + 1) – 3x2 + 2x – 5
c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)
\(=x^2+x-10\)
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1) 5(x-3) (x-7)-(5x+1) (x-2)= -8
2) x(x+1) (x+2)-(x+4) (3x-5)= 84-5x
3) (9x2-5) (x+3)-3x2(3x+9)=(x-5) (x+4)-x(x-11)
4) (x2-4x+16) (x+4)-x(x+1) (x+2)+3x2=0
5) (8x+2) (1-3x)+(6x-1) (4x-10)=-50
6) (x2+2x+4) (2-x)+x(x-3) (x+4)-x2+24=0
7) (\(\dfrac{x}{2}\)+3) (5-6x)+(12x-2) (\(\dfrac{x}{4}\)+3)=0
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
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1) (x2-4x+16) (x+4)-x(x+1) (x+2)+3x2=0
2) (8x+2) (1-3x)+(6x-1) (4x-10)=-50
3) (x2+2x+4) (2-x)+x(x-3) (x+4)-x2+24=0
4) (\(\dfrac{x}{2}\)x2+3) (5-6x)+(12x-2) (\(\dfrac{x}{4}\)x4+3)=0
1)(x2-4x+16)(x+4)-x(x+1)(x+2)+3x2=0
\(\Rightarrow\)(x3+64)-x(x2+2x+x+2)+3x2=0
\(\Rightarrow\)x3+64-x3-2x2-x2-2x+3x2=0
\(\Rightarrow\)-2x+64=0
\(\Rightarrow\)-2x=-64
\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)
\(\Rightarrow x=32\)
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2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50
\(\Rightarrow\)8x-24x2+2-6x+24x2-60x-4x+10=50
\(\Rightarrow\)-62x+12=50
\(\Rightarrow\)-62x=50-12
\(\Rightarrow\)-62x=38
\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)
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3)(x2+2x+4)(2-x)+x(x-3)(x+4)-x2+24=0
\(\Rightarrow\)8-x3+x(x2+4x-3x-12)-x2+24=0
\(\Rightarrow\)8-x3+x3+4x2-3x2-12x-x2+21=0
\(\Rightarrow\)-12x+29=0
\(\Rightarrow\)-12x=-29
\(\Rightarrow\)x=\(\dfrac{-29}{-12}=\dfrac{29}{12}\)
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Thực hiện phép tính:
a,(2x- 4)(x+9)
b,(x2 + 4x +3)(x-2)
c,(x-8)(x+8)
d, x2(7x-5)-7(x3- 4x+6)
e,(x2+2)(x2+x+1)
f,(x2+2)(x4-2x2+4)
g,(x-g)(x+9)
h,(x-2)(2x3-x2+1)+(x2+1)+(x2-2x2)(1-2)x
Dễ
Thế
Mà
Cũnhoir
Dc
Ạ
Chịu
Chắc
Phải
Ngu
Lamqs
Mới
Hỏi
Câu
Này
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a. x+1/x-2 - x/x+2 + 8/x2 -4
b. x-3/x+1 - x+2/x-1 + 8x/x2 -1
c. x+2/x2-2x + 2/x2+2x + 3x+2/x2-4
d. 4/x - 12/x2+3x + 5/x+3
a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
c: \(=\dfrac{x+2}{x\left(x-2\right)}+\dfrac{2}{x\left(x+2\right)}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x^2+2x+2x-4+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+7x-2}{x\left(x-2\right)\left(x+2\right)}\)
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a,
\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b,
\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)
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28 tháng 7 2021 lúc 9:29
ai giúp mình giải bài này với được k mình đang cần gấp ( xin cảm ơn)Bài 1: a,√3x+4−√2x+1√x+3b, √2x−5+√x+2√2x+1c, √x+4−√1−x√1−2xd, √x+95−√2x+4Bài 2:a,√x+4√x+45x+2b, √x2−2x+1+√x2+4x+44c, √x+2√x−1+√x−2√x−12d,√x−2+√2x−5+√x+2+3√2x−57√2Bài 3:a, x2−7x6√x+5−30
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ai giúp mình giải bài này với được k mình đang cần gấp ( xin cảm ơn)
Bài 1:
a,√3x+4−√2x+1=√x+3
b, √2x−5+√x+2=√2x+1
c, √x+4−√1−x=√1−2x
d, √x+9=5−√2x+4
Bài 2:
a,√x+4√x+4=5x+2
b, √x2−2x+1+√x2+4x+4=4
c, √x+2√x−1+√x−2√x−1=2
d,√x−2+√2x−5+√x+2+3√2x−5=7√2
Bài 3:
a, x2−7x=6√x+5−30
Tìm x biết:
a) (x+5).(2x+1)=0
b) x.(x+2)-3.(x+2)=0
c) 2x.(x-5)-x.(3+2x)=26
d) x2-10x-8x+16=0
e) x2-10x=25
f) 5x.(x-1)=x-1
g) 2.(x+5)-x2-5x=0
h) x2+5x-6=0
i) (2x-3)2-4.(x+1).(x-1)=49
j) x3+x2+x+1=0
k) x3-x2=4x2-8x+4
Mn ơi giúp em vs ạ,em cảm ơn trc ạ
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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