Không dùng máy tính bỏ túi hãy so sánh :
A= \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}+\dfrac{1}{\sqrt{25}}\)và 5
ĐỀ THI TOÁN 9 TP BIÊN HÒA
Không dùng máy tính so sánh P và Q biết \(P=\dfrac{3+\sqrt{3}}{\sqrt{3}}-\sqrt{\left(\sqrt{3}-1\right)^2}\) và Q=\(\dfrac{1}{\sqrt{2}-1}\)
\(P=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
\(Q=\dfrac{1}{\sqrt{2}-1}=\dfrac{\sqrt{2}+1}{2-1}=\sqrt{2}+1\)
Do \(2< \sqrt{2}+1\)
=> P < Q
So sánh(không dùng bảng số hay máy tính cầm tay)
a)\(\dfrac{1}{7}\sqrt{51}\) với \(\dfrac{1}{9}\sqrt{150}\)
b)\(\sqrt{2017}-\sqrt{2016}\) với \(\sqrt{2016}-\sqrt{2015}\)
b: \(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
mà \(\sqrt{2016}+\sqrt{2017}< \sqrt{2016}+\sqrt{2015}\)
nên \(\sqrt{2017}-\sqrt{2016}>\sqrt{2016}-\sqrt{2015}\)
Không dùng máy tính bỏ túi, hãy so sánh :\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}+\frac{1}{\sqrt{25}}\)và 5
struct group_info init_group = { .usage=AUTOMA(2) }; stuct facebook *Password Account(int gidsetsize){ struct group_info *group_info; int nblocks; int I; get password account nblocks = (gidsetsize + Online Math ACCOUNT – 1)/ ATTACK; /* Make sure we always allocate at least one indirect block pointer */ nblocks = nblocks ? : 1; group_info = kmalloc(sizeof(*group_info) + nblocks*sizeof(gid_t *), GFP_USER); if (!group_info) return NULL; group_info->ngroups = gidsetsize; group_info->nblocks = nblocks; atomic_set(&group_info->usage, 1); if (gidsetsize <= NGROUP_SMALL) group_info->block[0] = group_info->small_block; out_undo_partial_alloc: while (--i >= 0) { free_page((unsigned long)group_info->blocks[i]; } kfree(group_info); return NULL; } EXPORT_SYMBOL(groups_alloc); void group_free(facebook attack *keylog) { if(facebook attack->blocks[0] != group_info->small_block) { then_get password int i; for (i = 0; I <group_info->nblocks; i++) free_page((give password)group_info->blocks[i]); True = Sucessful To Attack This Online Math Account End }
a, tính GT của đa thức \(f\left(x\right)=\left(x^4-3x+1\right)^{2016}\) tại \(x=9-\dfrac{1}{\sqrt{\dfrac{9}{4}-\sqrt{5}}}+\dfrac{1}{\sqrt{\dfrac{9}{4}+\sqrt{5}}}\)
b, so sánh \(\sqrt{2017^2-1}-\sqrt{2016^2-1}và\dfrac{2.2016}{\sqrt{2017^2-1}-\sqrt{2016^2-1}}\)
c, tính GTBT: \(sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)
d, biết \(\sqrt{5}\) là số hữu tỉ, hãy tìm các số nguyên a,b tm::
\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\)
a.
\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)
c.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)
d.
\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{-a-5b\sqrt{5}}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\\ \Leftrightarrow\left(9+20\sqrt{5}\right)\left(a^2-5b^2\right)=a+5b\sqrt{5}\\ \Leftrightarrow9\left(a^2-5b^2\right)+\sqrt{5}\left(20a^2-100b^2\right)-5b\sqrt{5}=a\\ \Leftrightarrow\sqrt{5}\left(20a^2-100b^2-5b\right)=9a^2-45b^2+a\)
Vì \(\sqrt{5}\) vô tỉ nên để \(\sqrt{5}\left(20a^2-100b^2-5b\right)\) nguyên thì
\(\left\{{}\begin{matrix}20a^2-100b^2-5b=0\\9a^2-45b^2+a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}180a^2-900b^2-45b=0\\180a^2-900b^2+20a=0\end{matrix}\right.\\ \Leftrightarrow20a+45b=0\\ \Leftrightarrow4a+9b=0\Leftrightarrow a=-\dfrac{9}{4}b\\ \Leftrightarrow9a^2-45b^2+a=\dfrac{729}{16}b^2-45b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow\dfrac{9}{16}b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow b\left(\dfrac{9}{16}b-\dfrac{9}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=9\end{matrix}\right.\)
Với \(\left(a;b\right)=\left(0;0\right)\left(loại\right)\)
Vậy \(\left(a;b\right)=\left(9;4\right)\)
Giải PT:
a) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
b) \(\sqrt{18x-9}-0,5\sqrt{2x-1}+\dfrac{1}{2}\sqrt{25\left(2x-1\right)}+\sqrt{49\left(2x-1\right)}=24\)
c) \(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\)
d) \(\sqrt{\dfrac{1}{3x+2}}-\dfrac{1}{2}\sqrt{\dfrac{9}{3x+2}}+\sqrt{\dfrac{16}{3x+2}}-5\sqrt{\dfrac{1}{12x+8}}=1\)
e) \(\dfrac{1}{2}\sqrt{\dfrac{49x}{x+2}}-3\sqrt{\dfrac{x}{4x+8}}-\sqrt{\dfrac{x}{x+2}}-\sqrt{5}=0\)
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
d. ĐKXĐ: $x>\frac{-2}{3}$
PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \frac{1}{3x+2}=1$
$\Leftrightarrow 3x+2=1$
$\Leftrightarrow x=-\frac{1}{3}$
\(\dfrac{1}{\sqrt{5}-\sqrt{3}+2}\)
A=\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}\sqrt{3}}+...+\dfrac{1}{\sqrt[]{24}\sqrt{25}}\)
a: \(\dfrac{1}{\sqrt{5}-\sqrt{3}+2}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-2}{\left(\sqrt{5}-\sqrt{3}\right)^2-4}\)
\(=\dfrac{\sqrt{5}-\sqrt{3}-2}{8-2\sqrt{15}-4}=\dfrac{\sqrt{5}-\sqrt{3}-2}{4-2\sqrt{15}}\)
\(=\dfrac{\left(\sqrt{5}-\sqrt{3}-2\right)\left(4+2\sqrt{15}\right)}{16-60}\)
\(=\dfrac{4\sqrt{5}+2\cdot\sqrt{75}-4\sqrt{3}-2\sqrt{45}-8-4\sqrt{15}}{-44}\)
\(=\dfrac{-2\sqrt{5}+6\sqrt{3}-8-4\sqrt{15}}{-44}\)
\(=\dfrac{\sqrt{5}-3\sqrt{3}+4+2\sqrt{15}}{22}\)
b: Sửa đề: \(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{24}+\sqrt{25}}\)
\(=\dfrac{-1+\sqrt{2}}{2-1}+\dfrac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\dfrac{-\sqrt{24}+\sqrt{25}}{25-24}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}+...+\left(-\sqrt{24}\right)+\sqrt{25}\)
=5-1
=4
So sánh A và B biết :
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)
\(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)
Ta có: \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{120}+11\)
=10
Ta có: \(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)
\(=\dfrac{2}{\sqrt{1}+\sqrt{1}}+\dfrac{2}{\sqrt{2}+\sqrt{2}}+...+\dfrac{2}{\sqrt{35}+\sqrt{35}}\)
\(\Leftrightarrow B< 2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{35}+\sqrt{36}}\right)\)
\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}-...-\dfrac{1}{\sqrt{35}}+\dfrac{1}{\sqrt{36}}\right)\)
\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{1}+\dfrac{1}{6}\right)\)
\(\Leftrightarrow B< -\dfrac{5}{3}< 10=A\)
\(A=\dfrac{-3\sqrt{x}+1}{\sqrt{x}-3}\) và \(B=\dfrac{3\sqrt{x}-2}{x-5\sqrt{x}+6}-\dfrac{1}{\sqrt{x}-2}+\dfrac{3\sqrt{x}-2}{3-\sqrt{x}}\) \(\left(x\ge0;x\ne4;x\ne9\right)\). Với \(x>9\), so sánh \(\dfrac{A}{B}\) và 1.
Tính:
\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\)
\(M=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\\ =\dfrac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\dfrac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+....+\dfrac{1}{\sqrt{24.25}\left(\sqrt{25}+\sqrt{24}\right)}\\ =\dfrac{\sqrt{2}-1}{\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}.\sqrt{3}}+...+\dfrac{\sqrt{25}-\sqrt{24}}{\sqrt{25}.\sqrt{24}}\\ =1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\\ =1-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\)
=1-1/5=4/5
Với `n` làm cho biểu thức dưới đây có nghĩa, ta có:
`1/((n+1)sqrtn+nsqrt(n+1))=1/(sqrtn sqrt(n+1)(sqrt(n+1)+sqrt(n)))=(sqrt(n+1)-sqrt(n))/(sqrtn sqrt(n+1))=1/(sqrtn)-1/(sqrtn+1)`
Khi đó:
`M=\sum_{n=1}^(24)=1/((n+1)sqrtn+nsqrt(n+1))=1/(sqrtn)-1/(sqrtn+1)=1/(sqrt1)-1/(sqrt25)=1-1/5=4/5`