a) \(x^2-2xy-4z^2+y^2=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)=\left(6+4-2.45\right)\left(6+4+2.45\right)=-8000\)b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48=3\left(x^2+4x-21\right)+\left(x^2-8x+16\right)+48=4x^2+4x+1=\left(2x+1\right)^2=\left(2.0,5+1\right)^2=4\)

a: Ta có: \(x^2-2xy+y^2-4z^2\)

\(=\left(x-y\right)^2-\left(2z\right)^2\)

\(=\left(x-y-2z\right)\left(x-y+2z\right)\)

\(=\left(6+4-2\cdot45\right)\left(6+4+2\cdot45\right)\)

\(=-8000\)

b: Ta có: \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)

\(=3\left(x^2+4x-21\right)+\left(x-4\right)^2+48\)

\(=3x^2+12x-63+x^2-8x+16+48\)

\(=2x^2+4x+1\)

\(=2\cdot\dfrac{1}{4}+4\cdot\dfrac{1}{2}+1\)

\(=\dfrac{7}{2}\)

\(a,\)Thay \(x=-1;y=0,5\) vào \(p=x^3y-14y^3-6xy^2+y+2\)

\(\Rightarrow p=\left(-1\right)^3.0,5-14.0,5^3-6.\left(-1\right).0,5^2+0,5+2\)

\(=-0,5-1,75+1,5+0,5+2\)

\(=1,75\)

\(b,\)Thay \(x=0,2;y=-1,2\) vào \(q=15x^2y-5xy^2+7xy-21\)

\(\Rightarrow p=15.0,2^2.\left(-1,2\right)-5.0,2.\left(-1,2\right)^2+7.0,2.\left(-1,2\right)-21\)

\(=-0,72-1,44-1,68-21\)

\(=-24,84\)

1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)

\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)

\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)

\(=5-\dfrac{4}{2}\)

\(=5-2\)

\(=3\)

b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)

\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)

\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)

\(=0+0+0+2022\)

\(=2022\)

2) \(0,7^2\cdot x=0,49^2\)

\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)

\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)

\(\Rightarrow x=\left(0,7\right)^2\)

\(\Rightarrow x=0,49\)

b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)

\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)

\(\Rightarrow x=\left(-0,5\right)^5\)

\(\Rightarrow x=-\dfrac{1}{32}\)

2:

a: =>x*0,49=0,49^2

=>x=0,49

b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5

\(x^2-2xy-4z^2+y^2\)

\(=\left(x^2-2xy+y^2\right)-4z^2\)

\(=\left(x-y\right)^2-\left(2z\right)^2\)

\(=\left(x-y-2z\right)\left(x-y+2z\right)\)  ( 1 )

Thay vào bấm máy tính ta được ( 1 )=19

b) \(3\left(x-3\right)\left(x+7\right)-\left(x-4\right)^2\)

\(=\left(3x-9\right)\left(x+7\right)-\left(x^2-8x+16\right)\)

\(=3x^2+12-63-x^2+8x-16\)

\(=2x^2+20x-79\)

\(=2x^2+20x+50-129\)

\(=2\left(x+5\right)^2-129\)

Thay x vào 

Bài 2: Tính giá trị của biểu thức:
a) P= 1/3 x^2 y + xy^2 - xy + 1/2 xy^2 - 5xy - 1/3 x^2 y (1)

Tại x = 0,5; y = 1

Thay \(x=0,5 ; y=1\) vào biểu thức (1) , ta có :

P= \(\dfrac{1}{3} . 0,5^2.1+0,5.1^2-0,5.1+\dfrac{1}{2}. 0,5.1^2-5.0,5.1-\dfrac{1}{3}.0,5^2.1\)

P= \(=\dfrac{1}{12}+\dfrac{1}{2} -0,5+\dfrac{1}{4} -\dfrac{5}{2} - \dfrac{1}{12}\)

P= \(= \dfrac{-9}{4}\)

Vậy \(P =\dfrac{-9}{4}\)

Từ điều kiện �2+�2=2a2+b2=2, ta có (�+�)2−2��=2⇒��=12(�+�)2−1(a+b)2−2ab=2⇒ab=21​(a+b)2−1.

Đặt $x=a+b$.

Khi đó �=3�+12�2−1=12(�+3)2−112P=3x+21​x2−1=21​(x+3)2−211​.

Ta có (�+�)2≤2(�2+�2)⇒�2≤4⇒−2≤�≤2(a+b)2≤2(a2+b2)⇒x2≤4⇒−2≤x≤2.

Do đó �+3≥1⇒(�+3)2≥1⇒�≥−5x+3≥1⇒(x+3)2≥1⇒P≥−5.

Dấu bằng xảy ra khi $a=b=-1$.

Vậy giá trị nhỏ nhất của $P$ là $-5$.

Bài 1:

b) Ta có: \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}.\)

\(\Rightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{c^2}{16}.\)

\(\Rightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}\) và \(a^2-b^2+2c^2=108.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{a^2}{4}=4\Rightarrow a^2=16\Rightarrow\left[{}\begin{matrix}a=4\\a=-4\end{matrix}\right.\\\frac{b^2}{9}=4\Rightarrow b^2=36\Rightarrow\left[{}\begin{matrix}b=6\\b=-6\end{matrix}\right.\\\frac{c^2}{16}=4\Rightarrow c^2=64\Rightarrow\left[{}\begin{matrix}c=8\\c=-8\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(a;b;c\right)=\left(4;6;8\right),\left(-4;-6;-8\right).\)

Chúc bạn học tốt!

Mk sửa lại đề a

a) Tính (0,375-0,3+3/11+3/12)/(0,625-0,+5/11+5/12)

a) \(a:b=2\dfrac{2}{5}:\dfrac{4}{5}=\dfrac{12}{5}\cdot\dfrac{5}{4}=3:1\)

b) \(a:b=7.7:1.1=7:1\)

c) \(a:b=\dfrac{0.7\cdot100}{50}=\dfrac{70}{50}=\dfrac{7}{5}\)

d) \(a:b=\dfrac{3}{5}\cdot\dfrac{100}{120}=\dfrac{1}{2}\)

e) \(a:b=\dfrac{\dfrac{3}{2}\cdot60}{\dfrac{1}{2}}=3\cdot60=180:1\)

g) \(a=66\dfrac{2}{3}\%m=\dfrac{200}{3}\cdot\dfrac{1}{100}m=\dfrac{2}{3}m\)

\(b=0.5\%km=0.005km=5m\)

Do đó: \(a:b=\dfrac{2}{3}:5=\dfrac{2}{15}\)