\(130-12\left\{\left[18-\left(6:x+3\right)\right]-3\right\}=10\)
cac ban giai giai ho minh nhe
Những câu hỏi liên quan
Rut gon cac phan so sau
\(\frac{\left(-5\right)^2.\left(-6\right)^{11}.16^2-6^2.\left(-12\right)^6.15^2}{2.\left(-6\right)^{12}.10^4-81^2.960^3}\)
Cac ban giai nhanh va chi tiet cho mk nha
Minh xin cam on nhieu
Giai cac bpt sau
a,\(\left(x+1\right)\left(2x-2\right)-3>-5x-\left(2x+1\right)\left(3-x\right)\)
b,\(\left(x-3^{ }\right)^2+4\left(2-x\right)>\left(x+7\right)\)
a: \(\Leftrightarrow2x^2-2-3>-5x+\left(2x+1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2-5>-5x+2x^2-6x+x-3\)
\(\Leftrightarrow2x^2-5>2x^2-10x-3\)
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: \(\Leftrightarrow x^2-6x+9+8-4x>x+7\)
\(\Leftrightarrow x^2-10x+17-x-7>0\)
\(\Leftrightarrow x^2-11x+10>0\)
=>x>10 hoặc x<1
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a: ⇔2x2−2−3>−5x+(2x+1)(x−3)⇔2x2−2−3>−5x+(2x+1)(x−3)
⇔2x2−5>−5x+2x2−6x+x−3⇔2x2−5>−5x+2x2−6x+x−3
⇔2x2−5>2x2−10x−3⇔2x2−5>2x2−10x−3
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: ⇔x2−6x+9+8−4x>x+7⇔x2−6x+9+8−4x>x+7
⇔x2−10x+17−x−7>0⇔x2−10x+17−x−7>0
⇔x2−11x+10>0⇔x2−11x+10>0
=>x>10 hoặc x<1
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Giai cac bpt sau
a,\(\left(2x+3\right)\left(x+1\right)< 0\)
b,\(\left(4-x\right)\left(x+2\right)>0\)
a: (2x+3)(x+1)<0
=>2x+3 và x+1 khác dấu
=>x>-1 hoặc x<-3/2
b: (4-x)(x+2)>0
=>(x-4)(x+2)<0
=>-2<x<4
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a: (2x+3)(x+1)<0
=>2x+3 và x+1 khác dấu
=>x>-1 hoặc x<-3/2
b: (4-x)(x+2)>0
=>(x-4)(x+2)<0
=>-2<x<4
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\(\frac{2}{\left(x-1\right)\left(x-3\right)}\)+\(\frac{5}{\left(x-3\right)\left(x-8\right)}\)+\(\frac{12}{\left(x-3\right)\left(x-8\right)}\)-\(\frac{1}{x-20}\)=-\(\frac{3}{4}\)
Giup minh nhe cac ban :))☻ ☻
1 , \(c=\frac{10\frac{1}{3}.\left(26\frac{1}{3}-\frac{176}{7}\right)-\frac{12}{11}.\left(\frac{10}{3}-1,75\right)}{\frac{5}{\left(91-0,25\right).\frac{60}{11}-1}}\)
2 , \(A=\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,265+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
cac ban giai nhanh giup minh nha! ai nhanh minh k!
phan tich da thuc thanh nhan tu
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
ban nao giai ho mk bai nay voi
mk se tik cho
( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) - 24
= ( x2 + 7x + 10 ) ( x2 + 7x + 12 ) - 24
Đặt x2 + 7x + 10 = y
Ta có :
y2 + 2y - 24 = ( y - 4 ) ( y + 6 ) = ( x2 + 7x + 6 ) ( x2 + 7x + 16 )
= ( x + 1 ) ( x + 6 ) ( x2 + 7x + 16 )
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Đặt x2+7x+10=t
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=t\left(t+2\right)-24=t^2+2t-24\)
\(=\left(t^2+2t+1\right)-25=\left(t+1\right)^2-5^2=\left(t-4\right)\left(t+6\right)\)=(x2+7x+6)(x2+7x+16)
=(x2+x+6x+6)(x2+7x+16)=[x(x+1)+6(x+1)](x2+7x+16)=(x+1)(x+6)(x2+7x+16)
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\(B=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt: \(x^2+7x+10=t\)Khi đó B trở thành:
\(B=t\left(t+2\right)-24\)
\(=t^2+2t-24=\left(t-4\right)\left(t+6\right)\)
đến đây bạn thay trở lại
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a,\(\frac{2}{3}x-\frac{3}{2}\left(x-\frac{1}{2}\right)=\frac{5}{12}\)
b,\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)
AI GIAI DUOC MINH TICK CHO.NHO GIAI CHI TIET NHA
b,\(\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right):2=\frac{2013}{2015}:2\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow\)\(x+1=2015\)
\(\Rightarrow x=2014\)
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a, 2/3x -3/2.x-1/2x=5/12
x.(2/3-3/2-1/2)=5/12
x. -4/3=5/12
x=5/12:-4/3
x=-5/16
b,2/6+2/12+2/20+...+2/x.(x+1)=2013/2015
2/2.3+2/3.4+2/4.5+...+2/x.(x+1)=2013/2015
1/2(1-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)=2013/2015
1/2(1-1/x+1)=2013/2015
1-1/x+1=2013/2015 : 1/2
1-1/x+1=4206/2015
suy ra đề sai
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Tìm a và b để \(f\left(x\right)=x^4-3x^3+x^2+ax+b\) chia hết cho \(g\left(x\right)=x^3-3x+2\)
Cac ban giai giup minh voi minh cam on!!!
Dạng này bạn cứ đặt phép chia cho mình:) Rồi sau đó cho cái số dư = 0 để tìm a và b./.
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Rut gon
a)\(\left(x-4\right)\left(x+4\right)x-\left(x^2+1\right)\left(x^2-1\right)\)
b)\(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
c)\(x\left(x+\frac{1}{2}\right)-\left(2x-1\right)\left(x+\frac{3}{4}\right)\)
giai chi tiet giup minh nhe
a/-x4+x3-16x+1
b/-77
c/\(\frac{-4x^2+3}{4}\)
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