ĐKXĐ \(2x+1\ge0\Rightarrow x\ge-\frac{1}{2}\)

Khi đó |x² - 1| = 2x + 1

,=> \(\orbr{\begin{cases}x^2-1=2x+1\\x^2-1=-2x-1\end{cases}}\Rightarrow\orbr{\begin{cases}x^2-2x=2\\x^2+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2-2x+1=3\\x^2+2x+1=1\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=3\\\left(x+1\right)^2=1\end{cases}}\)

Nếu (x - 2)² = 3

=> \(\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{cases}}\)(tm)

Nếu (x + 1)² = 1

=> \(\orbr{\begin{cases}x+1=1\\x+1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\left(\text{loại}\right)\end{cases}}\Rightarrow x=0\)

a) \(A=\left(x-y\right)^2+\left(x+y\right)^2\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)

\(=x^2-2xy+y^2+x^2+2xy+y^2\)

\(=\left(x^2+x^2\right)-\left(2xy-2xy\right)+\left(y^2+y^2\right)\)

\(=2x^2+2y^2\)

\(=2.\left(x^2+y^2\right)\)

b) \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)

\(=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)

\(=4a^2+4ab+b^2-4a^2+4ab-b^2\)

\(=\left(4a^2-4a^2\right)+\left(4ab+4ab\right)+\left(b^2-b^2\right)\)

\(=8ab\)\

c) \(C=\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)

\(=x^2+2xy+y^2-x^2+2xy-y^2\)

\(=\left(x^2-x^2\right)+\left(2xy+2xy\right)+\left(y^2-y^2\right)\)

\(=4xy\)

d) \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+1-8x^2+24x-18+4\)

\(=\left(4x^2-8x^2\right)-\left(4x-24x\right)+\left(1-18+4\right)\)

\(=-4x^2+20x-13\)

\(=-4x^2+20x-25+12\)

\(=-\left(4x^2-20x+25\right)-8\)

\(=-\left[\left(2x\right)^2-2.4x.5+5^2\right]-8\)

\(=-\left(2x-5\right)^2-8\)

Có:     \(x^3+8+8\ge12x\)

VÀ:     \(y^3+27+27\ge27y\)             (LẦN LƯỢT ÁP DỤNG BĐT CAUCHY 3 SỐ)

VÀ:   \(\frac{x^3}{8}+\frac{y^3}{27}+1\ge\frac{xy}{2}\)

=>     \(\hept{\begin{cases}\frac{x^3}{8}+2\ge\frac{3x}{2}\\\frac{y^3}{27}+2\ge y\\\frac{x^3}{8}+\frac{y^3}{27}+1\ge\frac{xy}{2}\end{cases}}\)

CỘNG LẦN LƯỢT 3 BĐT TRÊN LẠI TA ĐƯỢC: 

=>     \(\frac{2x^3}{8}+\frac{2y^3}{27}+5\ge\frac{3x}{2}+y+\frac{xy}{2}\)

MÀ:     \(\frac{x}{2}+\frac{y}{3}+\frac{xy}{6}=3\)

=>      \(\frac{3x}{2}+y+\frac{xy}{2}=9\)

=>     \(\frac{2x^3}{8}+\frac{2y^3}{27}+5\ge9\)

=>       \(\frac{x^3}{8}+\frac{y^3}{27}\ge2\)

=>      \(\frac{27x^3+8y^3}{216}\ge2\)

=>       \(27x^3+8y^3\ge2.216=432\)

DẤU "=" XẢY RA <=>    \(x=2;y=3\)

VẬY P MIN = 432 <=>    x = 2;  y = 3.

a) \(5x\left(\frac{1}{5}x-2\right)+3\left(6-\frac{1}{3}x^2\right)=12\)

=> \(x^2-10x+18-x^2=12\)

=> -10x + 18 = 12

=> -10x = -6

=> -5x = -3

=> x = 3/5

b) 7x(x - 2) - 5(x - 1) = 7x² + 3

=> 7x² - 14x - 5x + 5 = 7x² + 3

=> 7x² - 14x - 5x + 5 - 7x² - 3 = 0

=> -19x + 2 = 0

=> -19x = -2

=> x = \(\frac{2}{19}\)

c) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11

=> 10x - 16 - 12x + 15 = 12x - 16 + 11

=> 10x - 16 - 12x + 15 - 12x + 16 - 11 = 0

=> (10x - 12x - 12x) + (-16 + 15 + 16 - 11) = 0

=> -14x + 4 = 0

=> -14x = -4

=> -7x = -2

=> x = 2/7

đề bài là tính nha các bn

mik ghi nhầm

Đặt \(Q=\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+...+\frac{1}{3^{99}}\)

\(\Rightarrow9Q=3+\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{97}}\)

\(\Rightarrow9Q-Q=\left(3+\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{97}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+...+\frac{1}{3^{99}}\right)\)

\(8Q=\frac{1}{3^{97}}-\frac{1}{3}\)

\(\Rightarrow Q=\frac{\frac{1}{3^{97}}-\frac{1}{3}}{8}\)

Vậy ...

Số cây khối lớp 5 trồng là :

250 : ( 7 - 5 ) x 7 = 875 ( cây )

Số cây khối lớp 4 trồng là :

875 - 250 = 625 ( cây )

Đáp số : Khối lớp 5 trồng 875 cây

             Khối lớp 4 trồng 625 cây

Khối lớp 5 trồng số cây là : 250 : ( 7 -5) . 7 = 875 ( cây)

Khối lớp 4 trồng số cây là : 875 - 250 = 625 ( cây)

Đáp số : ...

...

+) \(\left(x^2+1\right)\left(x+1\right)=4^y\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=2^{2y}\)

+) Do  \(x,y\inℕ\)nên ta có \(x^2+1=2^m\)và \(x+1=2^n\)với \(m+n=2y;m,n\inℕ\)

+) Lúc đó ta có: \(\orbr{\begin{cases}x^2+1⋮x+1\\x+1⋮x^2+1\end{cases}}\)

TH1: \(x^2+1⋮x+1\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)+2⋮x+1\)

\(\Leftrightarrow2⋮x+1\Leftrightarrow x\in\left\{0;1\right\}\)

TH2: \(x+1⋮x^2+1\Leftrightarrow x^2-1⋮x^2+1\Leftrightarrow2⋮x+1\)

\(\Leftrightarrow x\in\left\{0;1\right\}\)

* Nếu x = 0 thì \(4^y=1\Leftrightarrow y=0\)

* Nếu y = 0 thì \(4^y=4\Leftrightarrow y=1\)

Vậy \(\left(x;y\right)\in\left\{\left(0;0\right);\left(1;1\right)\right\}\)

quá dễ

1) 2-2+2-2=0;2/2*2/2=1;2/2+2/2=2;2-2/2+2=3;2+2-2+2=4;2+2+2/2=5;2*(2/2+2)=6;2+2+2+2=8;22/2-2=9;(22-2)/2=10

(còn số 7 không làm được)

2) a) \(52\cdot\frac{y}{78}=3380\)

\(52\cdot\frac{1}{78}\cdot y=3380\)

\(\frac{2}{3}\cdot y=3380\)

\(y=3380\div\frac{2}{3}=5070\)

b)\(55-y+33=76\)

\(88-y=76\)

\(y=88-76=12\)

3)\(3,54\cdot73+0,23\cdot25+3,54\cdot27+0,17\cdot25\)

\(=3,54\cdot\left(73+27\right)+25\cdot\left(0,27+0,13\right)\)

\(=3,54\cdot100+25\cdot0,4\)

\(=354+10=364\)

\(\frac{437^2-363^2}{537^2-463^2}=\frac{\left(437+363\right)\left(437-363\right)}{\left(537+463\right)\left(537-463\right)}=\frac{800.74}{1000.74}=\frac{8}{10}=\frac{4}{5}\)