a) đkxđ \(x\ge-1\)

pt đã cho tương đương với

\(x^2-x=2\left(\sqrt{x+1}-\sqrt{x^3+1}\right)\)

\(\Leftrightarrow x^2-x=2.\dfrac{x+1-\left(x^3+1\right)}{\sqrt{x+1}+\sqrt{x^3+1}}\)

\(\Leftrightarrow x\left(x-1\right)=2.\dfrac{x\left(1-x\right)}{\sqrt{x+1}+\sqrt{x^3+1}}\)

\(\Leftrightarrow x\left(x-1\right)\left[1+\dfrac{1}{\sqrt{x+1}+\sqrt{x^3+1}}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(nhận\right)\\1+\dfrac{1}{\sqrt{x+1}+\sqrt{x^3+1}}=0\left(vôlí\right)\end{matrix}\right.\)

Vậy pt đã cho có tâp nghiệm \(S=\left\{0;-1\right\}\)

\(x^2-x+2\sqrt[]{x^3+1}=2\sqrt[]{x+1}\) 

\(\Leftrightarrow2\sqrt[]{x^3+1}-2\sqrt[]{x+1}-\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{4}=0\)

\(\Leftrightarrow2\sqrt[]{x+1}\left(\sqrt[]{x^2-x+1}-1\right)-\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{4}=0\)

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

mà \(\left(x+\dfrac{1}{2}\right)^2\ge0,\forall x\inℝ\)

\(\left(1\right)\Leftrightarrow2\sqrt[]{x+1}\left(\sqrt[]{x^2-x+1}-1\right)-\dfrac{1}{4}\ge0\)

\(\Leftrightarrow\sqrt[]{x+1}\left(\sqrt[]{x^2-x+1}-1\right)\ge\dfrac{1}{8}\left(2\right)\)

Điều kiện xác định :

\(\left\{{}\begin{matrix}x+1\ge0\\\sqrt[]{x^2-x+1}-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\\sqrt[]{x^2-x+1}\ge1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-x+1\ge1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\left(x-1\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le0\cup x\ge1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\-1\le x\le0\end{matrix}\right.\)

BPT \(\left(2\right)\Leftrightarrow\left(x+1\right)\left(x^2-x+1-2\sqrt[]{x^2-x+1}-1\right)\ge\dfrac{1}{64}\)

\(\Leftrightarrow\left(x^2-x-2\sqrt[]{x^2-x+1}\right)\ge\dfrac{1}{64}\left(vì.x+1\ge0\right)\)

Đặt \(t=\sqrt[]{x^2-x+1}>0\)

\(BPT\Leftrightarrow t^2-2t-1-\dfrac{1}{64}\ge0\)

\(\Leftrightarrow t^2-2t-\dfrac{63}{64}\ge0\)

\(\Leftrightarrow t^2-2t+1-1-\dfrac{63}{64}\ge0\)

\(\Leftrightarrow\left(t-1\right)^2-\dfrac{127}{64}\ge0\)

\(\Leftrightarrow\left(t-1-\dfrac{\sqrt[]{127}}{8}\right)\left(t-1+\dfrac{\sqrt[]{127}}{8}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}t\ge1+\dfrac{\sqrt[]{127}}{8}\\t\le1-\dfrac{\sqrt[]{127}}{8}\end{matrix}\right.\)

\(\Leftrightarrow t\ge1+\dfrac{\sqrt[]{127}}{8}\)  \(\left(t>0;1-\dfrac{\sqrt[]{127}}{8}< 0\right)\)

\(\Leftrightarrow\sqrt[]{x^2-x+1}\ge1+\dfrac{\sqrt[]{127}}{8}\)

\(\Leftrightarrow x^2-x+1\ge\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2\)

mà \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\)

      \(\dfrac{3}{4}< \left(1+\dfrac{\sqrt[]{127}}{8}\right)^2\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\ge\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\ge\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}\le-\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}\\x-\dfrac{1}{2}\ge\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}+\dfrac{1}{2}\\x\ge\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}+\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}+\dfrac{1}{2}\) (so với đkxđ \(\left[{}\begin{matrix}x\ge1\\-1\le x\le0\end{matrix}\right.\))

\(\Leftrightarrow x=\sqrt[]{\left(1+\dfrac{\sqrt[]{127}}{8}\right)^2-\dfrac{3}{4}}+\dfrac{1}{2}\)

Ta có

MD//AB=> MD//AE

ME//AC=> ME//AD

=> ADME là hình bình hành (Tứ giác có các cặp cạnh đối // với nhau từng đôi một là hbh)=> ME=AD; MD=AE (cạnh đối hbh)

Ta có 

ME//AC \(\Rightarrow\dfrac{AE}{AB}=\dfrac{CM}{BC}\) (Talet trong tg) (1)

Ta có

MD//AB \(\Rightarrow\dfrac{AD}{AC}=\dfrac{BM}{BC}\) (Talet trong tg) (2)

Cộng 2 vế của (1) với (2)

\(\Rightarrow\dfrac{AE}{AB}+\dfrac{AD}{AC}=\dfrac{CM}{BC}+\dfrac{BM}{BC}=\dfrac{BC}{BC}=1\left(đpcm\right)\)

Lần sau những câu hỏi như vậy bạn dùng công thức trực quan để đặt câu hỏi nhé.

\(17.4^{x-3}-5.4^{x-4}=252\\ \Leftrightarrow\dfrac{17}{64}.4^x-\dfrac{5}{256}.4^x=252\\ \Leftrightarrow\dfrac{63}{256}.4^x=252\\ \Leftrightarrow4^x=1024\\ \Leftrightarrow4^x=4^5\\ \Leftrightarrow x=5\)

a/

S=1.2.(3-1)+2.3.(4-1)+3.4.(5-1)+...+99.100.(101-1)=

=1.2.3+2.3.4+3.4.5+...+99.100.101-(1.2+2.3+3.4+...+99.100)

Đặt

A=1.2.3+2.3.4+3.4.5+...+99.100.101

4A=1.2.3.4+2.3.4.4+3.4.5.4+...+99.100.101.4=

=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+...+99.100.101.(102-98)=

=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-...-98.99.100.101+99.100.101.102=

=99.100.101.102

=> A=99.100.101.102:4=99.25.100.102

Đặt 

B=1.2+2.3+3.4+...+99.100

3B=1.2.3+2.3.3+3.4.3+...+99.100.3=

=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)=

=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=

=99.100.101

=> B=99.100.101:3=33.100.101

=> S=A-B

Bạn tự tính nốt nhé

b/

Tổng trên có 51 số hạng

A=1+(2+2²)+(2³+2⁴)+...+(2⁴⁹+2⁵⁰)=

=1+2(1+2)+2³(1+2)+...+2⁴⁹(1+2)=

=1+3(2+2³+2⁵+...+2⁴⁹) => A:3 dư 1

Ta có

A=(1+2+2²)+(2³+2⁴+2⁵)+(2⁶+2⁷+2⁸)+...+(2⁴⁸+2⁴⁹+2⁵⁰)=

=7+2³(1+2+2²)+2⁶(1+2+2²)+...+2⁴⁸(1+2+2²)=

=7(1+2³+2⁶+...+2⁴⁸) chia hết cho 7

Ta có

A=1+2+2²+(2³+2⁴+2⁵+2⁶)+...+(2⁴⁷+2⁴⁸+2⁴⁹+2⁵⁰)=

=7+2³(1+2+2²+2³)+...+2⁴⁷(1+2+2²+2³)=

=7+15(2³+...+2⁴⁷)

=> A chia 15 dư 7

`#040911`

`175 - 4x^3 = 67`

`\Rightarrow 4x^3 = 175 - 67`

`\Rightarrow 4x^3 =108`

`\Rightarrow x^3 = 108 \div 4`

`\Rightarrow x^3 = 27`

`\Rightarrow x3 = 3^3`

`\Rightarrow x = 3`

Vậy, `x = 3.`

`#040911`

`x \div 61 = 9`

`\Rightarrow x = 9. 61`

`\Rightarrow x = 549`

Vậy, `x = 549.`

??????????????

Lời giải:
$(x^3+3x^2+3x+1):(x+1)=(x+1)^3:(x+1)=(x+1)^2$

Lời giải:

Đổi 4 dm = 0,4m

a. Diện tích mặt đáy thùng: $6,7\times 2=13,4$ (m²)

b. Diện tích xung quanh thùng: $2\times 0,4\times (6,7+2)=6,96$ (m²)

Diện tích quét sơn: $13,4+6,96=20,36$ (m²)

hùng = thùng nha

Lời giải:
a. 

Ta thấy: $(2x+4)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow A=(2x+4)^2-5\geq 0-5=-5$

Vậy $A_{\min}=-5$. Giá trị này đạt tại $2x+4=0\Leftrightarrow x=-2$

b.

Vì $(x+1)^2\geq 0$ với mọi $x$

$\Rightarrow -(x+1)^2\leq 0$ với mọi $x$

$\Rightarrow B=-(x+1)^2-5\leq 0-5=-5$

Vậy $B_{\max}=-5$. Giá trị này đạt tại $x+1=0\Leftrightarrow x=-1$