theo mình thì câu trên: dưới mẫu trong căn bỏ n^2 ra làm nhân tử chung xong đặt nhân tử chung của cả mẫu là n^2 . câu dưới thì mình k biết!!

\(\lim\dfrac{-3n+2}{n-\sqrt{4n+n^2}}=\lim\dfrac{\left(-3n+2\right)\left(n+\sqrt{4n+n^2}\right)}{\left(n-\sqrt{4n+n^2}\right)\left(n+\sqrt{4n+n^2}\right)}\)

\(=\lim\dfrac{\left(-3n+2\right)\left(n+\sqrt{4n+n^2}\right)}{-4n}=\lim\dfrac{n\left(-3+\dfrac{2}{n}\right)n\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4n}\)

\(=\lim n\dfrac{\left(-3+\dfrac{2}{n}\right)\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4}\)

Do \(\lim\left(n\right)=+\infty\)

\(\lim\dfrac{\left(-3+\dfrac{2}{n}\right)\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4}=\dfrac{\left(-3+0\right)\left(1+\sqrt{0+1}\right)}{-4}=\dfrac{3}{2}>0\)

\(\Rightarrow\lim n\dfrac{\left(-3+\dfrac{2}{n}\right)\left(1+\sqrt{\dfrac{4}{n}+1}\right)}{-4}=+\infty\)

\(\lim\left(\sqrt[3]{n^3+9n^2}-n\right)=\lim\dfrac{\left(\sqrt[3]{n^3+9n^2}-n\right)\left(\sqrt[3]{\left(n^3+9n^2\right)^2}+n\sqrt[3]{n^3+9n^2}+n^2\right)}{\sqrt[3]{\left(n^3+9n^2\right)}+n\sqrt[3]{n^3+9n^2}+n^2}\)

\(=\lim\dfrac{9n^2}{\sqrt[3]{\left(n^3+9n^2\right)^2}+n\sqrt[3]{n^3+9n^2}+n^2}\)

\(=\lim\dfrac{9n^2}{n^2\sqrt[3]{\left(1+\dfrac{9}{n}\right)^2}+n^2\sqrt[3]{1+\dfrac{9}{n}}+n^2}\)

\(=\lim\dfrac{9}{\sqrt[3]{\left(1+\dfrac{9}{n}\right)^2}+\sqrt[3]{1+\dfrac{9}{n}}+1}\)

\(=\dfrac{9}{\sqrt[3]{\left(1+0\right)^2}+\sqrt[3]{1+0}+1}=\dfrac{9}{3}=3\)

10.

\(H\left(x\right)=-5x^4+10x^3-15x+1\)

\(=-5x\left(x^3-2x^2+3\right)+1\)

\(=-5x.0+1\)

\(=1\)

9.

\(P\left(x\right)-Q\left(x\right)=\left(1-a\right)x^3+x^2+x-6\)

\(P\left(x\right)-Q\left(x\right)\) là đa thức bậc 3 khi và chỉ khi \(1-a\ne0\)

\(\Rightarrow a\ne1\)

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

\(y'=\dfrac{\left(-2x+2\right)\left(x-3\right)-\left(-x^2+2x+c\right)}{\left(x-3\right)^2}=\dfrac{-x^2+6x-6-c}{\left(x-3\right)^2}\)

\(\Rightarrow\) Cực đại và cực tiểu của hàm là nghiệm của: \(-x^2+6x-6-c=0\) (1)

\(\Delta'=9-\left(6+c\right)>0\Rightarrow c< 3\)

Gọi \(x_1;x_2\) là 2 nghiệm của (1) \(\Rightarrow\left\{{}\begin{matrix}-x_1^2+6x_1-6=c\\-x_2^2+6x_2-6=c\end{matrix}\right.\)

\(\Rightarrow m-M=\dfrac{-x_1^2+2x_1+c}{x_1-3}-\dfrac{-x_2^2+2x_2+c}{x_2-3}=4\)

\(\Leftrightarrow\dfrac{-2x_1^2+8x_1-6}{x_1-3}-\dfrac{-2x_2^2+8x_2-6}{x_2-3}=4\)

\(\Leftrightarrow2\left(1-x_1\right)-2\left(1-x_2\right)=4\)

\(\Leftrightarrow x_2-x_1=2\)

Kết hợp với Viet: \(\left\{{}\begin{matrix}x_2-x_1=2\\x_1+x_2=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=2\\x_2=4\end{matrix}\right.\)

\(\Rightarrow c=2\)

Có 1 giá trị nguyên

Ta có:

     (2 - 3x)(x + 8) = (3x - 2)(3 - 5x)

⇔ (2 - 3x)(x + 8) - (3x - 2)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8) + (2 - 3x)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8 + 3 - 5x) = 0

⇔ (2 - 3x)(11 - 4x) = 0

⇔ 2 - 3x = 0 hay 11 - 4x = 0

⇔ 2 = 3x hay 11 = 4x

⇔ x = \(\dfrac{2}{3}\) hay x = \(\dfrac{11}{4}\)

Vậy tập nghiệm của pt S = \(\left\{\dfrac{2}{3};\dfrac{11}{4}\right\}\)

khum bt mới lớp 5 hoi à(k10)

<=> (2-3x ) (x+8) + (2-3x ) (3-5x)=0
<=> (2-3x ) ( x+8 +  3-5x ) =0 
<=> (2-3x ) ( 11 - 4x ) = 0
 => 2-3x  =0 hoặc 11-4x =0  
       3x = 2            4x =11
         x = 2/3         x    = 11/4

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x-2}+1}{\sqrt[]{x+3}-2}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{x-2}+1\right)\left(\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1\right)\left(\sqrt[]{x+3}+2\right)}{\left(\sqrt[]{x+3}-2\right)\left(\sqrt[]{x+3}+2\right)\left(\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(\sqrt[]{x+3}+2\right)}{\left(x-1\right)\left(\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{x+3}+2}{\sqrt[3]{\left(x-2\right)^2}-\sqrt[3]{x-2}+1}\)

\(=\dfrac{\sqrt[]{1+3}+2}{\sqrt[3]{\left(1-2\right)^2}-\sqrt[3]{1-2}+1}=\dfrac{4}{3}\)

mai bạn tách ra nha để vậy hơi nhiều

c1: theo ct: \(I=\dfrac{U}{R}\)=>U tỉ lệ thuận I =>I càng lớn thì U càng lớn

C2(bn làm đúng)

C3: \(=>Umax=Imax.R=40.\dfrac{250}{1000}=10V\)=>chọn C

c4: R1 nt(R2//R3) =>U2=U3 mà R2=R3=>I2=I3

\(=>I1=I2+I3=>I2=I3=\dfrac{I1}{2}\)

C5: R1 nt R2

mà \(I1=2A,I2=1,5A\)=>chọn I2\(=>I1=I2=Im=1,5A=>Umax=\left(R1+R2\right).1,5=90V\)

C6: R1//R2

\(=>U1=I1R1=30V,U2=I2R2=15V\)=.chọn U2

C7\(=>\dfrac{1}{RTd}=\dfrac{1}{R1}+\dfrac{1}{R2}+\dfrac{1}{R3}=>Rtd=6\left(om\right)\)

C8-\(=>I=\dfrac{U}{\dfrac{R1R2}{R1+R2}}=0,9A\)

\(=>I1=\dfrac{U}{R1}=\dfrac{12}{20}=0,6A=>I2=0,3A\)

C9-\(=>U3=\left(\dfrac{U1}{R1}\right)R3=8V=>Um=U1+U2+U3=....\)

(thay số vào)

C10\(=>\dfrac{1}{Rtd}=\dfrac{1}{R1}+\dfrac{1}{R2}+\dfrac{1}{R3}=>Rtd=......\)(thay số)

C11: các bóng đèn như nhau nên mắc vào chung 1 nguồn điện nối tiếp sẽ hoạt động với đúng cường độ dòng điện định mức nên các bóng đều sáng bth=>chọn B

C12 \(\dfrac{1}{Rtd}=\dfrac{1}{R1}+\dfrac{1}{R2}\)=>chọn D

c13\(=>R=\dfrac{U}{I}=\dfrac{6}{0,3}=20\left(om\right)\)

c14 R1 nt R2

\(R1=\dfrac{3}{0,3}=10\left(om\right),R2=\dfrac{6}{0,5}=12\left(om\right)=>I1=I2=\dfrac{11}{R1+R2}=0,5A=>I1>I\left(đm1\right),I2=I\left(đm2\right)\)

=>đèn 1 sáng mạnh hơn bth có thể hỏng , đèn 2 sáng bth

c15.\(=>\dfrac{R1}{R2}=\dfrac{S2}{S1}=>\dfrac{R1}{6}=\dfrac{1}{3}=>R1=2\left(om\right)\)

c16.\(=>l=\dfrac{RS}{p}=\dfrac{\left(\dfrac{U}{I}\right)S}{p}=\dfrac{\left(\dfrac{220}{5}\right).2.10^{-6}}{0,4.10^{-6}}=220m\)

c17.=>\(S'=3S,=>l'=\dfrac{1}{3}l\)

\(=>\dfrac{R}{R'}=\dfrac{\dfrac{pl}{S}}{\dfrac{pl'}{S'}}=\dfrac{S'.l}{S.l'}=\dfrac{3S.l}{S.\dfrac{1}{3}.l}=9=>R=9R'=>R'=\dfrac{R}{9}=1\left(om\right)\)

c18.chọn dây dẫn R3 có l3=l2,S3=S1,chùng chất liệu đồng

\(=>\dfrac{R1}{R3}=\dfrac{l1}{l3}=>\dfrac{1,7}{R3}=\dfrac{100}{200}=>R3=3,4\left(om\right)\)

\(=>\dfrac{R2}{R3}=\dfrac{S3}{S2}=>\dfrac{17}{3,4}=\dfrac{10^{-6}}{S2}=>S2=2.10^{-7}m^2\)\(=0,2mm^2\)

c19 \(l1=8l2,S1=2S2\)

\(=>\dfrac{R1}{R2}=\dfrac{\dfrac{pl1}{S1}}{\dfrac{.pl2}{S2}}=\dfrac{S2.l1}{S1.l2}=\dfrac{S2.8l2}{2S2.l2}=4=>R1=4R2\)

c20.\(=>R=\dfrac{0,9}{15}=0,06\left(om\right)\)(đáp án đề sai)

c21\(=>l=\dfrac{RS}{p}=\dfrac{10.10^{-7}}{0,4.10^{-6}}=2,5m\)

c22\(=>R=\dfrac{pl}{S}=\dfrac{6.1;7.10^{-8}}{3,14.\left(\dfrac{0,0012}{2}\right)^2}=0,09\left(om\right)\)

\(\lim\dfrac{3^n+2.6^n}{6^{n-1}+5.4^n}=\lim\dfrac{6^n\left[\left(\dfrac{3}{6}\right)^n+2\right]}{6^n\left[\dfrac{1}{6}+5\left(\dfrac{4}{6}\right)^n\right]}=\lim\dfrac{\left(\dfrac{3}{6}\right)^n+2}{\dfrac{1}{6}+5\left(\dfrac{4}{6}\right)^n}=\dfrac{0+2}{\dfrac{1}{6}+0}=12\)

\(\lim\left(\sqrt{n^2+9}-n\right)=\lim\dfrac{\left(\sqrt{n^2+9}-n\right)\left(\sqrt{n^2+9}+n\right)}{\sqrt{n^2+9}+n}=\lim\dfrac{9}{\sqrt{n^2+9}+n}\)

\(=\lim\dfrac{n\left(\dfrac{9}{n}\right)}{n\left(\sqrt{1+\dfrac{9}{n^2}}+1\right)}=\lim\dfrac{\dfrac{9}{n}}{\sqrt{1+\dfrac{9}{n^2}}+1}=\dfrac{0}{1+1}=0\)

\(\lim\dfrac{\sqrt{15+9n^2}-3}{5-n}=\lim\dfrac{n\sqrt{\dfrac{15}{n^2}+9}-3}{5-n}=\lim\dfrac{n\left(\sqrt{\dfrac{15}{n^2}+9}-\dfrac{3}{n}\right)}{n\left(\dfrac{5}{n}-1\right)}\)

\(=\lim\dfrac{\sqrt{\dfrac{15}{n^2}+9}-\dfrac{3}{n}}{\dfrac{5}{n}-1}=\dfrac{\sqrt{9}-0}{0-1}=-3\)

\(I=\int\dfrac{2}{2+5sinxcosx}dx=\int\dfrac{2sec^2x}{2sec^2x+5tanx}dx\\ =\int\dfrac{2sec^2x}{2tan^2x+5tanx+2}dx\)

We substitute :

\(u=tanx,du=sec^2xdx\\ I=\int\dfrac{2}{2u^2+5u+2}du\\ =\int\dfrac{2}{2\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{8}}du\\ =\int\dfrac{1}{\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{16}}du\\ \)

Then, 

\(t=u+\dfrac{5}{4}\\I=\int\dfrac{1}{t^2-\dfrac{9}{16}}dt\\ =\int\dfrac{\dfrac{2}{3}}{t-\dfrac{3}{4}}-\dfrac{\dfrac{2}{3}}{t+\dfrac{3}{4}}dt\)

Finally,

\(I=\dfrac{2}{3}ln\left(\left|\dfrac{t-\dfrac{3}{4}}{t+\dfrac{3}{4}}\right|\right)+C=\dfrac{2}{3}ln\left(\left|\dfrac{tanx+\dfrac{1}{2}}{tanx+2}\right|\right)+C\)