a) \(A=\left(\dfrac{\sqrt[]{a}}{\sqrt[]{a}+\sqrt[]{b}}+\dfrac{a}{b-a}\right):\left(\dfrac{\sqrt[]{a}}{\sqrt[]{a}+\sqrt[]{b}}-\dfrac{a}{a+b+2\sqrt[]{ab}}\right)\left(a;b>0;a\ne\pm b\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt[]{a}\left(\sqrt[]{b}-\sqrt[]{a}\right)+a}{\left(\sqrt[]{a}+\sqrt[]{b}\right)\left(\sqrt[]{b}-\sqrt[]{a}\right)}\right):\left(\dfrac{\sqrt[]{a}\left(\sqrt[]{a}+\sqrt[]{b}\right)-a}{\left(\sqrt[]{a}+\sqrt[]{b}\right)^2}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt[]{ab}-a+a}{\left(\sqrt[]{a}+\sqrt[]{b}\right)\left(\sqrt[]{b}-\sqrt[]{a}\right)}\right):\left(\dfrac{a+\sqrt[]{ab}-a}{\left(\sqrt[]{a}+\sqrt[]{b}\right)^2}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt[]{ab}}{\left(\sqrt[]{a}+\sqrt[]{b}\right)\left(\sqrt[]{b}-\sqrt[]{a}\right)}\right):\left(\dfrac{\sqrt[]{ab}}{\left(\sqrt[]{a}+\sqrt[]{b}\right)^2}\right)\)
\(\Leftrightarrow A=\dfrac{\sqrt[]{ab}}{\left(\sqrt[]{a}+\sqrt[]{b}\right)\left(\sqrt[]{b}-\sqrt[]{a}\right)}.\dfrac{\left(\sqrt[]{a}+\sqrt[]{b}\right)^2}{\sqrt[]{ab}}\)
\(\Leftrightarrow A=\dfrac{a+b+2\sqrt[]{ab}}{b-a}\)
b) Với \(a=7-4\sqrt[]{3};b=7+4\sqrt[]{3}\)
\(\Leftrightarrow A=\dfrac{7-4\sqrt[]{3}+7+4\sqrt[]{3}+2\sqrt[]{\left(7-4\sqrt[]{3}\right)\left(7+4\sqrt[]{3}\right)}}{7+4\sqrt[]{3}-\left(7-4\sqrt[]{3}\right)}\)
\(\Leftrightarrow A=\dfrac{14+2\sqrt[]{\left[\left(7^2-\left(4\sqrt[]{3}\right)^2\right)\right]}}{7+4\sqrt[]{3}-7+4\sqrt[]{3}}\)
\(\Leftrightarrow A=\dfrac{14+2\sqrt[]{\left(49-48\right)}}{8\sqrt[]{3}}=\dfrac{16}{8\sqrt[]{3}}=\dfrac{2}{\sqrt[]{3}}=\dfrac{2\sqrt[]{3}}{3}\)
a, 4x-√3(3x-1)=3x-1
b, 3√2(x+1)=1-4x
c, 6x-3√4(2x-1)=2x+1
a) \(4x-\sqrt[]{3\left(3x-1\right)}=3x-1\)
\(\Leftrightarrow\sqrt[]{3\left(3x-1\right)}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\3\left(3x-1\right)=\left(x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\9x-3=x^2+2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\left(a\right)\\x^2-7x+4=0\left(1\right)\end{matrix}\right.\)
Giải \(pt\left(1\right):\)
\(\Delta=49-16=33\Rightarrow\sqrt[]{\Delta}=\sqrt[]{33}\)
Phương trình (1) có 2 nghiệm phân biệt
\(\left[{}\begin{matrix}x=\dfrac{7+\sqrt[]{33}}{2}\\x=\dfrac{7-\sqrt[]{33}}{2}\end{matrix}\right.\) (thỏa \(\left(a\right)\))
tìm x:
a,32:(3xX-2)=8
b,75:(X-18)=25
c,(15-6xX)x243=729
d,4x(X-12)+9=17
e,20-2x(X+4)=4
a) 32 : (3.x - 2) = 8
3x - 2 = 32 : 8
3x - 2 = 4
3x = 4 + 2
3x = 6
x = 6 : 3
x = 2
b) 75 : (x - 18) = 25
x - 18 = 75 : 25
x - 18 = 3
x = 3 + 18
x = 21
c) (15 - 6.x) . 243 = 729
15 - 6x = 729 : 243
15 - 6x = 3
6x = 15 - 3
6x = 12
x = 12 : 6
x = 2
d) 4.(x - 12) + 9 = 17
4(x - 12) = 17 - 9
4(x - 12) = 8
x - 12 = 8 : 4
x - 12 = 2
x = 2 + 12
x = 14
e) 20 - 2.(x + 4) = 4
2(x + 4) = 20 - 4
2(x + 4) = 16
x + 4 = 16 : 2
x + 4 = 8
x = 8 : 2
x = 4
`32: ( 3xx x -2)=8`
`3xx x-2=32:8`
`3xx x-2=4`
`3 xx x=4+2`
`3xx x=6`
`x=6:3`
`x=2`
__
`75 : (x-18) =25`
`x-18=75:25`
`x-18= 3`
`x=3+18`
`x=21`
__
`(15-6 xx x ) xx 243 =729`
`15-6 xx x = 729 : 243`
`15-6 xx x = 3`
`6 xx x=15-3`
`6 xx x=12`
`x=12:6`
`x=2`
__
`4 xx (x-12)+9=17`
`4 xx (x-12)=17-9`
`4 xx (x-12)= 8`
`x-12=8:4`
`x-12=2`
`x=2+12`
`x=14`
__
`20-2xx(x+4)=4`
`2xx(x+4)=20-4`
`2xx(x+4)=16`
`x+4=16:2`
`x+4=8`
`x=8-4`
`x=4`
\(a,32:\left(3\times X-2\right)=8\\ 3\times X-2=32:8=4\\ 3\times X=4+2=6\\ X=\dfrac{6}{3}=2\\ Vậy:X=2\\ ---\\ b,75:\left(X-18\right)=25\\ X-18=75:25=3\\ X=3+18=21\\ Vậy:X=21\\ ---\\ c,\left(15-6\times X\right)\times243=729\\ 15-6\times X=729:243=3\\ 6\times X=15-3=12\\ X=\dfrac{12}{6}=2\\ Vậy:X=2\\ ---\\ d,4\times\left(X-12\right)+9=17\\ 4\times\left(X-12\right)=17-9=8\\ X-12=8:4=2\\ X=2+12=14\\ Vậy:X=14\\ ---\\ e,20-2\times\left(X+4+4\right)=4\\ 2\times\left(X+4\right)=20-4=16\\ X+4=16:2=8\\ X=8-4=4\\ Vậy:X=4\)
tìm GTLN ( hoặc nhỏ nhất) của đa thức sau
(x-1)^2+3 1-x^2
\(a,\) Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2+3\ge3\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy \(GTNN\) của đa thức là \(3\) khi \(x=1.\)
\(b,\) Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow-x^2\le0\forall x\)
\(\Rightarrow1-x^2\le1\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
#\(Toru\)
Đặt A = (x - 1)² + 3
Do (x - 1)² ≥ 0 với mọi x ∈ R
⇒ (x - 1)² + 3 ≥ 3 với mọi x ∈ R
Vậy GTNN của A là 3 khi x = 1
---------------
Đặt B = 1 - x² = -x² + 1
Do x² ≥ 0 với mọi x ∈ R
⇒ -x² ≤ 0 với mọi x ∈ R
⇒ -x² + 1 ≤ 1 với mọi x ∈ R
Vậy GTLN của B là 1 khi x = 0
giúp nhanh và chi tiết cho mik ạ
\(1) ( 2x+1)^2 + 2(2x+1) + 1\)
\(=\left(2x+1\right)^2+2\cdot\left(2x+1\right)\cdot1+1^2\)
\(=\left(2x+1+1\right)^2\)
\(=\left(2x+2\right)^2\)
\(2) (3x-2y)^2+4(3x-2y)+4\)
\(=\left(3x-2y\right)^2+2\cdot\left(3x-2y\right)\cdot2+2^2\)
\(=\left(3x-2y+2\right)^2\)
#\(Toru\)
`# \text {04th5.}`
`1)`
`(2x + 1)^2 + 2(2x + 1) + 1`
`= 4x^2 + 4x + 1 + 4x + 2 + 1`
`= 4x^2 + 8 + 4`
`2)`
`(3x - 2y)^2 + 4(3x - 2y) + 4`
`= 9x^2 - 12xy + 4y^2 + 12x - 8y + 4`
Một đội công nhân sửa đường trong 3 ngày.Ngày đầu sửa đc 2/13 đoạn đường,ngày thứ 2 sửa đc 13/4 đoạn đường.Hỏi ngày thứ 3 còn phải sửa bao nhiêu đoạn đường? Mong mn giúp e ạ
Cho hàm số y= ( \(\dfrac{-3}{2}\) m +5 )x -6
a) tìm m để hàm số trên đồng biến
b) Vẽ đồ thị khi m=2
\(a,HSĐB\Leftrightarrow-\dfrac{3}{2}m+5>0\Leftrightarrow m>\dfrac{10}{3}\\ b,m=2\Rightarrow y=2x-6\\ Chọn.3.điểm:A\left(0;-6\right);B\left(2;-2\right);C\left(3;0\right)\)
Anh chọn điểm em tự vẽ đồ thị hi
a) Hàm số đồng biến khi:
-3m/2 + 5 > 0
⇔ -3m/2 > -5
⇔ m < 10/3
b) m = 2
⇔ y = 2x - 6
Cho x = 0 thì y = -6 ⇒ A(0; -6)
y = 0 thì x = 3 ⇒ B(3; 0)
*) Đồ thị:
tính
\(3\dfrac{1}{5}\times\dfrac{10}{11}+1\dfrac{2}{11}=\)
\(5\dfrac{1}{3}:1\dfrac{2}{3}-1\dfrac{1}{5}=\)
mik cần gấp mn 1
\(3\dfrac{1}{5}\times\dfrac{10}{11}+1\dfrac{2}{11}\\ =\dfrac{16}{5}\times\dfrac{10}{11}+\dfrac{13}{11}\\ =\dfrac{16\times10}{5\times11}+\dfrac{13}{11}\\ =\dfrac{160}{55}+\dfrac{13}{11}\\ =\dfrac{32}{11}+\dfrac{13}{11}\\ =\dfrac{32+13}{11}\\ =\dfrac{45}{11}\)
\(5\dfrac{1}{3}:1\dfrac{2}{3}-1\dfrac{1}{5}\\ =\dfrac{16}{3}:\dfrac{5}{3}-\dfrac{6}{5}\\ =\dfrac{16}{3}\times\dfrac{3}{5}-\dfrac{6}{5}\\ =\dfrac{16\times3}{3\times5}-\dfrac{6}{5}\\ =\dfrac{48}{15}-\dfrac{6}{5}\\ =\dfrac{16}{5}-\dfrac{6}{5}\\ =\dfrac{16-6}{5}\\ =\dfrac{10}{5}\\ =2\)
\(3\dfrac{1}{5}\times\dfrac{10}{11}+1\dfrac{2}{11}\\ =\dfrac{3\times5+1}{5}\times\dfrac{10}{11}+\dfrac{1\times11+2}{11}\\ =\dfrac{16}{5}\times\dfrac{10}{11}+\dfrac{13}{11}\\ =\dfrac{160}{55}+\dfrac{13}{11}\\ =\dfrac{32}{11}+\dfrac{13}{11}\\ =\dfrac{45}{11}\)
\(5\dfrac{1}{3}:1\dfrac{2}{3}-1\dfrac{1}{5}\\ =\dfrac{5\times3+1}{3}:\dfrac{1\times3+2}{3}-\dfrac{1\times5+1}{5}\\ =\dfrac{16}{3}:\dfrac{5}{3}-\dfrac{6}{5}\\ =\dfrac{16}{3}\times\dfrac{3}{5}-\dfrac{6}{5}\\ =\dfrac{16}{5}-\dfrac{6}{5}\\ =\dfrac{10}{5}=2\)
Cho A= \(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}\)và B= \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\)
a) rút gọn B
b) Cho x>0. so sánh A với 3
\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)
\(=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
`b,` Tớ tính mãi ko ra, xl cậu nha=')
phân tích đa thức thành nhân tử (x+-1)(x+2)(x+3)+(x+6)-28
Sửa đề: \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-28\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)-28\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-28\)
\(=\left(x^2+5x\right)^2-64\)
\(=\left(x^2+5x+8\right)\left(x^2+5x-8\right)\)